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Compute $limlimits_xto infty sumlimits_n=1^infty frac1n(n+x)$
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I want to compute $$lim_xto infty sum_n=1^infty frac1n(n+x).$$
Can I do as follow? Consider the measurable space $(mathbb N,mathcal P(mathbb N),mu)$ where $mu(A)=#A$. Then,
$$sum_n=1^infty frac1n(n+x)=int_mathbb Nfrac1n(n+x)dmu(n).$$
Suppose $|x|geq 1$. Then
$$left|frac1n(n+x)right|leq frac1n(n+1)in L^1(mathbb N),$$
and thus, using DCT, we finally obtain $$lim_xto infty sum_n=1^infty frac1n(n+x)=sum_n=1^infty lim_xto infty frac1n(n+x)=0.$$
Does it work ?
real-analysis sequences-and-series measure-theory summation
edited Jul 22 at 13:32
Did
242k23208443
asked Jul 22 at 13:11
user380364
927214
 |Â
show 6 more comments
up vote
8
down vote
favorite
I want to compute $$lim_xto infty sum_n=1^infty frac1n(n+x).$$
Can I do as follow? Consider the measurable space $(mathbb N,mathcal P(mathbb N),mu)$ where $mu(A)=#A$. Then,
$$sum_n=1^infty frac1n(n+x)=int_mathbb Nfrac1n(n+x)dmu(n).$$
Suppose $|x|geq 1$. Then
$$left|frac1n(n+x)right|leq frac1n(n+1)in L^1(mathbb N),$$
and thus, using DCT, we finally obtain $$lim_xto infty sum_n=1^infty frac1n(n+x)=sum_n=1^infty lim_xto infty frac1n(n+x)=0.$$
Does it work ?
real-analysis sequences-and-series measure-theory summation
edited Jul 22 at 13:32
Did
242k23208443
asked Jul 22 at 13:11
user380364
927214
Don’t know if your method works, but I would use M test to prove uniform convergence and then take limit termwise.
– Szeto
Jul 22 at 13:19
1
The dominated convergence theorem is for sequences of functions. In your case you have a limit over the real numbers. You can apply your argument to deduce that for every monotonic sequence $x_mtoinfty$ you have $lim_mtoinftysum_n=1^inftyfrac1n(n+x_m)=0$. Then you need to deduce from that that $lim_xtoinftysum_n=1^inftyfrac1n(n+x)=0$.
– user574889
Jul 22 at 13:23
1
It is important to remember that the dominated convergence theorem is for sequences because of this
– user574889
Jul 22 at 13:27
1
This is a good approach. Use it to show that the limit when $xtoinfty$, $x$ integer, is zero, then extend the result to $xtoinfty$, $x$ real, by monotonicity of the sum with respect to $x$.
– Did
Jul 22 at 13:30
1
@user380364 Monotonicity is not needed. What is important is that on $mathbbR$ a limit $lim_xto a$ exists (as a unique limit) if and only if it exists for all sequences $x_nto a$. This is ultimately due to $mathbbR$ being first countable.
– user574889
Jul 22 at 14:13
 |Â
show 6 more comments
up vote
8
down vote
favorite
up vote
8
down vote
favorite
I want to compute $$lim_xto infty sum_n=1^infty frac1n(n+x).$$
Can I do as follow? Consider the measurable space $(mathbb N,mathcal P(mathbb N),mu)$ where $mu(A)=#A$. Then,
$$sum_n=1^infty frac1n(n+x)=int_mathbb Nfrac1n(n+x)dmu(n).$$
Suppose $|x|geq 1$. Then
$$left|frac1n(n+x)right|leq frac1n(n+1)in L^1(mathbb N),$$
and thus, using DCT, we finally obtain $$lim_xto infty sum_n=1^infty frac1n(n+x)=sum_n=1^infty lim_xto infty frac1n(n+x)=0.$$
Does it work ?
real-analysis sequences-and-series measure-theory summation
edited Jul 22 at 13:32
Did
242k23208443
asked Jul 22 at 13:11
user380364
927214
I want to compute $$lim_xto infty sum_n=1^infty frac1n(n+x).$$
Can I do as follow? Consider the measurable space $(mathbb N,mathcal P(mathbb N),mu)$ where $mu(A)=#A$. Then,
$$sum_n=1^infty frac1n(n+x)=int_mathbb Nfrac1n(n+x)dmu(n).$$
Suppose $|x|geq 1$. Then
$$left|frac1n(n+x)right|leq frac1n(n+1)in L^1(mathbb N),$$
and thus, using DCT, we finally obtain $$lim_xto infty sum_n=1^infty frac1n(n+x)=sum_n=1^infty lim_xto infty frac1n(n+x)=0.$$
Does it work ?
real-analysis sequences-and-series measure-theory summation
edited Jul 22 at 13:32
Did
242k23208443
asked Jul 22 at 13:11
user380364
927214
edited Jul 22 at 13:32
Did
242k23208443
edited Jul 22 at 13:32
Did
242k23208443
edited Jul 22 at 13:32
Did
242k23208443
242k23208443
asked Jul 22 at 13:11
user380364
927214
asked Jul 22 at 13:11
user380364
927214
asked Jul 22 at 13:11
user380364
927214
927214
Don’t know if your method works, but I would use M test to prove uniform convergence and then take limit termwise.
– Szeto
Jul 22 at 13:19
1
The dominated convergence theorem is for sequences of functions. In your case you have a limit over the real numbers. You can apply your argument to deduce that for every monotonic sequence $x_mtoinfty$ you have $lim_mtoinftysum_n=1^inftyfrac1n(n+x_m)=0$. Then you need to deduce from that that $lim_xtoinftysum_n=1^inftyfrac1n(n+x)=0$.
– user574889
Jul 22 at 13:23
1
It is important to remember that the dominated convergence theorem is for sequences because of this
– user574889
Jul 22 at 13:27
1
This is a good approach. Use it to show that the limit when $xtoinfty$, $x$ integer, is zero, then extend the result to $xtoinfty$, $x$ real, by monotonicity of the sum with respect to $x$.
– Did
Jul 22 at 13:30
1
@user380364 Monotonicity is not needed. What is important is that on $mathbbR$ a limit $lim_xto a$ exists (as a unique limit) if and only if it exists for all sequences $x_nto a$. This is ultimately due to $mathbbR$ being first countable.
– user574889
Jul 22 at 14:13
 |Â
show 6 more comments
Don’t know if your method works, but I would use M test to prove uniform convergence and then take limit termwise.
– Szeto
Jul 22 at 13:19
1
The dominated convergence theorem is for sequences of functions. In your case you have a limit over the real numbers. You can apply your argument to deduce that for every monotonic sequence $x_mtoinfty$ you have $lim_mtoinftysum_n=1^inftyfrac1n(n+x_m)=0$. Then you need to deduce from that that $lim_xtoinftysum_n=1^inftyfrac1n(n+x)=0$.
– user574889
Jul 22 at 13:23
1
It is important to remember that the dominated convergence theorem is for sequences because of this
– user574889
Jul 22 at 13:27
1
This is a good approach. Use it to show that the limit when $xtoinfty$, $x$ integer, is zero, then extend the result to $xtoinfty$, $x$ real, by monotonicity of the sum with respect to $x$.
– Did
Jul 22 at 13:30
1
@user380364 Monotonicity is not needed. What is important is that on $mathbbR$ a limit $lim_xto a$ exists (as a unique limit) if and only if it exists for all sequences $x_nto a$. This is ultimately due to $mathbbR$ being first countable.
– user574889
Jul 22 at 14:13
Don’t know if your method works, but I would use M test to prove uniform convergence and then take limit termwise.
– Szeto
Jul 22 at 13:19
Don’t know if your method works, but I would use M test to prove uniform convergence and then take limit termwise.
– Szeto
Jul 22 at 13:19
1
1
The dominated convergence theorem is for sequences of functions. In your case you have a limit over the real numbers. You can apply your argument to deduce that for every monotonic sequence $x_mtoinfty$ you have $lim_mtoinftysum_n=1^inftyfrac1n(n+x_m)=0$. Then you need to deduce from that that $lim_xtoinftysum_n=1^inftyfrac1n(n+x)=0$.
– user574889
Jul 22 at 13:23
The dominated convergence theorem is for sequences of functions. In your case you have a limit over the real numbers. You can apply your argument to deduce that for every monotonic sequence $x_mtoinfty$ you have $lim_mtoinftysum_n=1^inftyfrac1n(n+x_m)=0$. Then you need to deduce from that that $lim_xtoinftysum_n=1^inftyfrac1n(n+x)=0$.
– user574889
Jul 22 at 13:23
1
1
It is important to remember that the dominated convergence theorem is for sequences because of this
– user574889
Jul 22 at 13:27
It is important to remember that the dominated convergence theorem is for sequences because of this
– user574889
Jul 22 at 13:27
1
1
This is a good approach. Use it to show that the limit when $xtoinfty$, $x$ integer, is zero, then extend the result to $xtoinfty$, $x$ real, by monotonicity of the sum with respect to $x$.
– Did
Jul 22 at 13:30
This is a good approach. Use it to show that the limit when $xtoinfty$, $x$ integer, is zero, then extend the result to $xtoinfty$, $x$ real, by monotonicity of the sum with respect to $x$.
– Did
Jul 22 at 13:30
1
1
@user380364 Monotonicity is not needed. What is important is that on $mathbbR$ a limit $lim_xto a$ exists (as a unique limit) if and only if it exists for all sequences $x_nto a$. This is ultimately due to $mathbbR$ being first countable.
– user574889
Jul 22 at 14:13
@user380364 Monotonicity is not needed. What is important is that on $mathbbR$ a limit $lim_xto a$ exists (as a unique limit) if and only if it exists for all sequences $x_nto a$. This is ultimately due to $mathbbR$ being first countable.
– user574889
Jul 22 at 14:13
 |Â
show 6 more comments
4 Answers
4
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oldest
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up vote
3
down vote
Yes, your proof is fine (at least modulo cactus's comment: You really do need to reduce to a sequence before you can apply DCT. This is no big deal, people apply DCT to $lim_xtoinfty$ all the time, because it's clear that $lim_xto inftyI_x=I$ if and only if $lim_ntoinftyI_x_n=I$ for every sequence $x_n$ with $x_ntoinfty$.)
Of course there's a very simple more elementary argument here, which I write out because it leads to Something Interesting at the bottom:
Elementary Argument: Let $epsilon>0$. Choose $N$ so $$sum_n=N+1^inftyfrac1n^2<epsilon.$$Now for every $x>0$ we have
$$sum_1^inftyfrac1n(n+x)<epsilon+sum_1^Nfrac1n(n+x);$$since $N$ is fixed it follows that $sum_1^infty<2epsilon$ if $x$ is large enough.
Something Interesting: Exercise 1 Generalize the argument above to give an elementary proof of DCT for coutning measure on $Bbb N$.
Exercise 2 Show that DCT for a general measure space follows from Egoroff's Theorem, by an argument analogous to the argument above.
I like the proof of DCT via Egoroff - at least to me it gives a much better picture of "why it's really true" then the proof from Fatou's Lemmma that you see in all the books.
answered Jul 22 at 13:42
David C. Ullrich
54.1k33481
But why we always pass by a sequence to apply DCT ? Let's call "Corollary of DCT" the fact that $lim_xto infty int_mathbb Rf(t,x)dt=int_mathbb Rlim_xtoinfty f(x,t)dt$ always work ? (I made a proof in the comment of my question). I don't see in what we must pass with a sequence. Yes DCT with Egoroff's theorem is more intuitive since in fact all sequence $(f_n)$ that converge p.p. to $f$ converge almost uniformly.
– user380364
Jul 22 at 13:51
Officially we need to pass to sequences because DCT applies to sequences. In fact people do apply your "Corollary to DCT" all the time without mentioning it, because it's more or less obvious how to reduce to sequences in that case. But as cactus points out, there are other notions of limit that cannot be reduced to sequences, and for which in fact DCT fails.
– David C. Ullrich
Jul 22 at 14:02
add a comment |Â
up vote
2
down vote
As pointed out in comments, the OP's approach is largely correct; it's only flaw is that the DCT is usually stated for sequences of functions, not a continuum of them, so the proof needs an extra step along the lines of
$$lim_xtoinftysum1over n(x+x)lelim_xtoinftysum1over n(x+lfloor xrfloor)=lim_mtoinftysum1over n(n+m)$$
(where $m$ is understood as an integer variable).
So as an exercise in understanding the DCT, this is a nice example.
For what it's worth, here is another alternative approach to showing the limit is $0$: By the Arithmetic-Geometric Mean inequality, we have $n+xge2sqrtnx$, and thus
$$sum1over n(n+x)le1over2sqrt xsum1over n^3/2$$
answered Jul 22 at 13:48
Barry Cipra
56.5k652118
(+1) This is clear and concise, and embeds an elementary way forward using the AM-GM. Well done!
– Mark Viola
Jul 22 at 15:49
add a comment |Â
up vote
1
down vote
In the same spirit as A. Pongrácz in his/her answer, after partial fraction decomposition
$$S_x=sum_n=1^infty frac1n(n+x)=fracH_xx$$ Using the asymptotics of harmonic numbers, we then have
$$S_x=fracgamma +log left(xright)x+frac12 x^2-frac112
x^3+Oleft(frac1x^5right)$$ which seems to be quite good even for small values of $x$.
Forexample $$S_10=frac738125200approx 0.29289683$$ while the above expansion gives
$$frac5912000+frac110 (gamma +log (10))approx 0.29289674$$
answered Jul 22 at 14:06
Claude Leibovici
111k1055126
add a comment |Â
up vote
-1
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I am not sure if I understood what you wrote. But I am quite sure that you do not need the measure you defined, or in fact any complicated calculus.
The function $f(x):= sumlimits_n=1^infty frac1n(n+x)$ is clearly strictly monotone decreasing and non-negative. So the limit exists, and in fact $limlimits_xrightarrow infty f(x)$ is the limit of the sequence $limlimits_mrightarrow infty f(m)$ where $m$ runs through the positive integers.
So we can restrict our attention to positive integers $x$. Then $frac1n(n+x) = frac1x (frac1n- frac1n+x)$. The sum $sumlimits_n=1^infty frac1n- frac1n+x$ is a telescopic sum, its value is $sumlimits_n=1^x frac1n approx log x$. So $f(x)approx fraclog xx$, which tends to zero as $x$ tends to infinity.
answered Jul 22 at 13:27
A. Pongrácz
2,109221
It's not my question. I don't want to compute this limit. I just want to make a comparaison with measure theory.
– user380364
Jul 22 at 13:28
1
I do not think it can be effectively used to solve this problem. If anything, it seems to complicate matters rather than simplify them.
– A. Pongrácz
Jul 22 at 13:30
On the contrary, it is a very simple approach to see that the terms converge to zero and then go looking for a dominant sequence and then fill in the issue of the real limit by monotonicity. DCT is often taught only later on, but the concept is simply enough "you are actually allowed to do what you have always wanted to do if you check this condition".
– Phira
Jul 22 at 13:39
In fact the calculus you use here is much more complicated than necessary.
– David C. Ullrich
Jul 22 at 14:03
We can make it more general without restricting $x$ to integers. Starting from your answer, I built mine. By the way, $+1$ for your to compensate a downvote which is not justified (at least for me). Cheers.
– Claude Leibovici
Jul 22 at 14:11
 |Â
show 1 more comment
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Yes, your proof is fine (at least modulo cactus's comment: You really do need to reduce to a sequence before you can apply DCT. This is no big deal, people apply DCT to $lim_xtoinfty$ all the time, because it's clear that $lim_xto inftyI_x=I$ if and only if $lim_ntoinftyI_x_n=I$ for every sequence $x_n$ with $x_ntoinfty$.)
Of course there's a very simple more elementary argument here, which I write out because it leads to Something Interesting at the bottom:
Elementary Argument: Let $epsilon>0$. Choose $N$ so $$sum_n=N+1^inftyfrac1n^2<epsilon.$$Now for every $x>0$ we have
$$sum_1^inftyfrac1n(n+x)<epsilon+sum_1^Nfrac1n(n+x);$$since $N$ is fixed it follows that $sum_1^infty<2epsilon$ if $x$ is large enough.
Something Interesting: Exercise 1 Generalize the argument above to give an elementary proof of DCT for coutning measure on $Bbb N$.
Exercise 2 Show that DCT for a general measure space follows from Egoroff's Theorem, by an argument analogous to the argument above.
I like the proof of DCT via Egoroff - at least to me it gives a much better picture of "why it's really true" then the proof from Fatou's Lemmma that you see in all the books.
answered Jul 22 at 13:42
David C. Ullrich
54.1k33481
But why we always pass by a sequence to apply DCT ? Let's call "Corollary of DCT" the fact that $lim_xto infty int_mathbb Rf(t,x)dt=int_mathbb Rlim_xtoinfty f(x,t)dt$ always work ? (I made a proof in the comment of my question). I don't see in what we must pass with a sequence. Yes DCT with Egoroff's theorem is more intuitive since in fact all sequence $(f_n)$ that converge p.p. to $f$ converge almost uniformly.
– user380364
Jul 22 at 13:51
Officially we need to pass to sequences because DCT applies to sequences. In fact people do apply your "Corollary to DCT" all the time without mentioning it, because it's more or less obvious how to reduce to sequences in that case. But as cactus points out, there are other notions of limit that cannot be reduced to sequences, and for which in fact DCT fails.
– David C. Ullrich
Jul 22 at 14:02
add a comment |Â
up vote
3
down vote
Yes, your proof is fine (at least modulo cactus's comment: You really do need to reduce to a sequence before you can apply DCT. This is no big deal, people apply DCT to $lim_xtoinfty$ all the time, because it's clear that $lim_xto inftyI_x=I$ if and only if $lim_ntoinftyI_x_n=I$ for every sequence $x_n$ with $x_ntoinfty$.)
Of course there's a very simple more elementary argument here, which I write out because it leads to Something Interesting at the bottom:
Elementary Argument: Let $epsilon>0$. Choose $N$ so $$sum_n=N+1^inftyfrac1n^2<epsilon.$$Now for every $x>0$ we have
$$sum_1^inftyfrac1n(n+x)<epsilon+sum_1^Nfrac1n(n+x);$$since $N$ is fixed it follows that $sum_1^infty<2epsilon$ if $x$ is large enough.
Something Interesting: Exercise 1 Generalize the argument above to give an elementary proof of DCT for coutning measure on $Bbb N$.
Exercise 2 Show that DCT for a general measure space follows from Egoroff's Theorem, by an argument analogous to the argument above.
I like the proof of DCT via Egoroff - at least to me it gives a much better picture of "why it's really true" then the proof from Fatou's Lemmma that you see in all the books.
answered Jul 22 at 13:42
David C. Ullrich
54.1k33481
But why we always pass by a sequence to apply DCT ? Let's call "Corollary of DCT" the fact that $lim_xto infty int_mathbb Rf(t,x)dt=int_mathbb Rlim_xtoinfty f(x,t)dt$ always work ? (I made a proof in the comment of my question). I don't see in what we must pass with a sequence. Yes DCT with Egoroff's theorem is more intuitive since in fact all sequence $(f_n)$ that converge p.p. to $f$ converge almost uniformly.
– user380364
Jul 22 at 13:51
Officially we need to pass to sequences because DCT applies to sequences. In fact people do apply your "Corollary to DCT" all the time without mentioning it, because it's more or less obvious how to reduce to sequences in that case. But as cactus points out, there are other notions of limit that cannot be reduced to sequences, and for which in fact DCT fails.
– David C. Ullrich
Jul 22 at 14:02
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Yes, your proof is fine (at least modulo cactus's comment: You really do need to reduce to a sequence before you can apply DCT. This is no big deal, people apply DCT to $lim_xtoinfty$ all the time, because it's clear that $lim_xto inftyI_x=I$ if and only if $lim_ntoinftyI_x_n=I$ for every sequence $x_n$ with $x_ntoinfty$.)
Of course there's a very simple more elementary argument here, which I write out because it leads to Something Interesting at the bottom:
Elementary Argument: Let $epsilon>0$. Choose $N$ so $$sum_n=N+1^inftyfrac1n^2<epsilon.$$Now for every $x>0$ we have
$$sum_1^inftyfrac1n(n+x)<epsilon+sum_1^Nfrac1n(n+x);$$since $N$ is fixed it follows that $sum_1^infty<2epsilon$ if $x$ is large enough.
Something Interesting: Exercise 1 Generalize the argument above to give an elementary proof of DCT for coutning measure on $Bbb N$.
Exercise 2 Show that DCT for a general measure space follows from Egoroff's Theorem, by an argument analogous to the argument above.
I like the proof of DCT via Egoroff - at least to me it gives a much better picture of "why it's really true" then the proof from Fatou's Lemmma that you see in all the books.
answered Jul 22 at 13:42
David C. Ullrich
54.1k33481
Yes, your proof is fine (at least modulo cactus's comment: You really do need to reduce to a sequence before you can apply DCT. This is no big deal, people apply DCT to $lim_xtoinfty$ all the time, because it's clear that $lim_xto inftyI_x=I$ if and only if $lim_ntoinftyI_x_n=I$ for every sequence $x_n$ with $x_ntoinfty$.)
Of course there's a very simple more elementary argument here, which I write out because it leads to Something Interesting at the bottom:
Elementary Argument: Let $epsilon>0$. Choose $N$ so $$sum_n=N+1^inftyfrac1n^2<epsilon.$$Now for every $x>0$ we have
$$sum_1^inftyfrac1n(n+x)<epsilon+sum_1^Nfrac1n(n+x);$$since $N$ is fixed it follows that $sum_1^infty<2epsilon$ if $x$ is large enough.
Something Interesting: Exercise 1 Generalize the argument above to give an elementary proof of DCT for coutning measure on $Bbb N$.
Exercise 2 Show that DCT for a general measure space follows from Egoroff's Theorem, by an argument analogous to the argument above.
I like the proof of DCT via Egoroff - at least to me it gives a much better picture of "why it's really true" then the proof from Fatou's Lemmma that you see in all the books.
answered Jul 22 at 13:42
David C. Ullrich
54.1k33481
answered Jul 22 at 13:42
David C. Ullrich
54.1k33481
answered Jul 22 at 13:42
David C. Ullrich
54.1k33481
answered Jul 22 at 13:42
David C. Ullrich
54.1k33481
54.1k33481
But why we always pass by a sequence to apply DCT ? Let's call "Corollary of DCT" the fact that $lim_xto infty int_mathbb Rf(t,x)dt=int_mathbb Rlim_xtoinfty f(x,t)dt$ always work ? (I made a proof in the comment of my question). I don't see in what we must pass with a sequence. Yes DCT with Egoroff's theorem is more intuitive since in fact all sequence $(f_n)$ that converge p.p. to $f$ converge almost uniformly.
– user380364
Jul 22 at 13:51
Officially we need to pass to sequences because DCT applies to sequences. In fact people do apply your "Corollary to DCT" all the time without mentioning it, because it's more or less obvious how to reduce to sequences in that case. But as cactus points out, there are other notions of limit that cannot be reduced to sequences, and for which in fact DCT fails.
– David C. Ullrich
Jul 22 at 14:02
add a comment |Â
But why we always pass by a sequence to apply DCT ? Let's call "Corollary of DCT" the fact that $lim_xto infty int_mathbb Rf(t,x)dt=int_mathbb Rlim_xtoinfty f(x,t)dt$ always work ? (I made a proof in the comment of my question). I don't see in what we must pass with a sequence. Yes DCT with Egoroff's theorem is more intuitive since in fact all sequence $(f_n)$ that converge p.p. to $f$ converge almost uniformly.
– user380364
Jul 22 at 13:51
Officially we need to pass to sequences because DCT applies to sequences. In fact people do apply your "Corollary to DCT" all the time without mentioning it, because it's more or less obvious how to reduce to sequences in that case. But as cactus points out, there are other notions of limit that cannot be reduced to sequences, and for which in fact DCT fails.
– David C. Ullrich
Jul 22 at 14:02
But why we always pass by a sequence to apply DCT ? Let's call "Corollary of DCT" the fact that $lim_xto infty int_mathbb Rf(t,x)dt=int_mathbb Rlim_xtoinfty f(x,t)dt$ always work ? (I made a proof in the comment of my question). I don't see in what we must pass with a sequence. Yes DCT with Egoroff's theorem is more intuitive since in fact all sequence $(f_n)$ that converge p.p. to $f$ converge almost uniformly.
– user380364
Jul 22 at 13:51
But why we always pass by a sequence to apply DCT ? Let's call "Corollary of DCT" the fact that $lim_xto infty int_mathbb Rf(t,x)dt=int_mathbb Rlim_xtoinfty f(x,t)dt$ always work ? (I made a proof in the comment of my question). I don't see in what we must pass with a sequence. Yes DCT with Egoroff's theorem is more intuitive since in fact all sequence $(f_n)$ that converge p.p. to $f$ converge almost uniformly.
– user380364
Jul 22 at 13:51
Officially we need to pass to sequences because DCT applies to sequences. In fact people do apply your "Corollary to DCT" all the time without mentioning it, because it's more or less obvious how to reduce to sequences in that case. But as cactus points out, there are other notions of limit that cannot be reduced to sequences, and for which in fact DCT fails.
– David C. Ullrich
Jul 22 at 14:02
Officially we need to pass to sequences because DCT applies to sequences. In fact people do apply your "Corollary to DCT" all the time without mentioning it, because it's more or less obvious how to reduce to sequences in that case. But as cactus points out, there are other notions of limit that cannot be reduced to sequences, and for which in fact DCT fails.
– David C. Ullrich
Jul 22 at 14:02
add a comment |Â
up vote
2
down vote
As pointed out in comments, the OP's approach is largely correct; it's only flaw is that the DCT is usually stated for sequences of functions, not a continuum of them, so the proof needs an extra step along the lines of
$$lim_xtoinftysum1over n(x+x)lelim_xtoinftysum1over n(x+lfloor xrfloor)=lim_mtoinftysum1over n(n+m)$$
(where $m$ is understood as an integer variable).
So as an exercise in understanding the DCT, this is a nice example.
For what it's worth, here is another alternative approach to showing the limit is $0$: By the Arithmetic-Geometric Mean inequality, we have $n+xge2sqrtnx$, and thus
$$sum1over n(n+x)le1over2sqrt xsum1over n^3/2$$
answered Jul 22 at 13:48
Barry Cipra
56.5k652118
(+1) This is clear and concise, and embeds an elementary way forward using the AM-GM. Well done!
– Mark Viola
Jul 22 at 15:49
add a comment |Â
up vote
2
down vote
As pointed out in comments, the OP's approach is largely correct; it's only flaw is that the DCT is usually stated for sequences of functions, not a continuum of them, so the proof needs an extra step along the lines of
$$lim_xtoinftysum1over n(x+x)lelim_xtoinftysum1over n(x+lfloor xrfloor)=lim_mtoinftysum1over n(n+m)$$
(where $m$ is understood as an integer variable).
So as an exercise in understanding the DCT, this is a nice example.
For what it's worth, here is another alternative approach to showing the limit is $0$: By the Arithmetic-Geometric Mean inequality, we have $n+xge2sqrtnx$, and thus
$$sum1over n(n+x)le1over2sqrt xsum1over n^3/2$$
answered Jul 22 at 13:48
Barry Cipra
56.5k652118
(+1) This is clear and concise, and embeds an elementary way forward using the AM-GM. Well done!
– Mark Viola
Jul 22 at 15:49
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As pointed out in comments, the OP's approach is largely correct; it's only flaw is that the DCT is usually stated for sequences of functions, not a continuum of them, so the proof needs an extra step along the lines of
$$lim_xtoinftysum1over n(x+x)lelim_xtoinftysum1over n(x+lfloor xrfloor)=lim_mtoinftysum1over n(n+m)$$
(where $m$ is understood as an integer variable).
So as an exercise in understanding the DCT, this is a nice example.
For what it's worth, here is another alternative approach to showing the limit is $0$: By the Arithmetic-Geometric Mean inequality, we have $n+xge2sqrtnx$, and thus
$$sum1over n(n+x)le1over2sqrt xsum1over n^3/2$$
answered Jul 22 at 13:48
Barry Cipra
56.5k652118
As pointed out in comments, the OP's approach is largely correct; it's only flaw is that the DCT is usually stated for sequences of functions, not a continuum of them, so the proof needs an extra step along the lines of
$$lim_xtoinftysum1over n(x+x)lelim_xtoinftysum1over n(x+lfloor xrfloor)=lim_mtoinftysum1over n(n+m)$$
(where $m$ is understood as an integer variable).
So as an exercise in understanding the DCT, this is a nice example.
For what it's worth, here is another alternative approach to showing the limit is $0$: By the Arithmetic-Geometric Mean inequality, we have $n+xge2sqrtnx$, and thus
$$sum1over n(n+x)le1over2sqrt xsum1over n^3/2$$
answered Jul 22 at 13:48
Barry Cipra
56.5k652118
edited Jul 22 at 13:55
answered Jul 22 at 13:48
Barry Cipra
56.5k652118
answered Jul 22 at 13:48
Barry Cipra
56.5k652118
answered Jul 22 at 13:48
Barry Cipra
56.5k652118
56.5k652118
(+1) This is clear and concise, and embeds an elementary way forward using the AM-GM. Well done!
– Mark Viola
Jul 22 at 15:49
add a comment |Â
(+1) This is clear and concise, and embeds an elementary way forward using the AM-GM. Well done!
– Mark Viola
Jul 22 at 15:49
(+1) This is clear and concise, and embeds an elementary way forward using the AM-GM. Well done!
– Mark Viola
Jul 22 at 15:49
(+1) This is clear and concise, and embeds an elementary way forward using the AM-GM. Well done!
– Mark Viola
Jul 22 at 15:49
add a comment |Â
up vote
1
down vote
In the same spirit as A. Pongrácz in his/her answer, after partial fraction decomposition
$$S_x=sum_n=1^infty frac1n(n+x)=fracH_xx$$ Using the asymptotics of harmonic numbers, we then have
$$S_x=fracgamma +log left(xright)x+frac12 x^2-frac112
x^3+Oleft(frac1x^5right)$$ which seems to be quite good even for small values of $x$.
Forexample $$S_10=frac738125200approx 0.29289683$$ while the above expansion gives
$$frac5912000+frac110 (gamma +log (10))approx 0.29289674$$
answered Jul 22 at 14:06
Claude Leibovici
111k1055126
add a comment |Â
up vote
1
down vote
In the same spirit as A. Pongrácz in his/her answer, after partial fraction decomposition
$$S_x=sum_n=1^infty frac1n(n+x)=fracH_xx$$ Using the asymptotics of harmonic numbers, we then have
$$S_x=fracgamma +log left(xright)x+frac12 x^2-frac112
x^3+Oleft(frac1x^5right)$$ which seems to be quite good even for small values of $x$.
Forexample $$S_10=frac738125200approx 0.29289683$$ while the above expansion gives
$$frac5912000+frac110 (gamma +log (10))approx 0.29289674$$
answered Jul 22 at 14:06
Claude Leibovici
111k1055126
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In the same spirit as A. Pongrácz in his/her answer, after partial fraction decomposition
$$S_x=sum_n=1^infty frac1n(n+x)=fracH_xx$$ Using the asymptotics of harmonic numbers, we then have
$$S_x=fracgamma +log left(xright)x+frac12 x^2-frac112
x^3+Oleft(frac1x^5right)$$ which seems to be quite good even for small values of $x$.
Forexample $$S_10=frac738125200approx 0.29289683$$ while the above expansion gives
$$frac5912000+frac110 (gamma +log (10))approx 0.29289674$$
answered Jul 22 at 14:06
Claude Leibovici
111k1055126
In the same spirit as A. Pongrácz in his/her answer, after partial fraction decomposition
$$S_x=sum_n=1^infty frac1n(n+x)=fracH_xx$$ Using the asymptotics of harmonic numbers, we then have
$$S_x=fracgamma +log left(xright)x+frac12 x^2-frac112
x^3+Oleft(frac1x^5right)$$ which seems to be quite good even for small values of $x$.
Forexample $$S_10=frac738125200approx 0.29289683$$ while the above expansion gives
$$frac5912000+frac110 (gamma +log (10))approx 0.29289674$$
answered Jul 22 at 14:06
Claude Leibovici
111k1055126
answered Jul 22 at 14:06
Claude Leibovici
111k1055126
answered Jul 22 at 14:06
Claude Leibovici
111k1055126
answered Jul 22 at 14:06
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
up vote
-1
down vote
I am not sure if I understood what you wrote. But I am quite sure that you do not need the measure you defined, or in fact any complicated calculus.
The function $f(x):= sumlimits_n=1^infty frac1n(n+x)$ is clearly strictly monotone decreasing and non-negative. So the limit exists, and in fact $limlimits_xrightarrow infty f(x)$ is the limit of the sequence $limlimits_mrightarrow infty f(m)$ where $m$ runs through the positive integers.
So we can restrict our attention to positive integers $x$. Then $frac1n(n+x) = frac1x (frac1n- frac1n+x)$. The sum $sumlimits_n=1^infty frac1n- frac1n+x$ is a telescopic sum, its value is $sumlimits_n=1^x frac1n approx log x$. So $f(x)approx fraclog xx$, which tends to zero as $x$ tends to infinity.
answered Jul 22 at 13:27
A. Pongrácz
2,109221
It's not my question. I don't want to compute this limit. I just want to make a comparaison with measure theory.
– user380364
Jul 22 at 13:28
1
I do not think it can be effectively used to solve this problem. If anything, it seems to complicate matters rather than simplify them.
– A. Pongrácz
Jul 22 at 13:30
On the contrary, it is a very simple approach to see that the terms converge to zero and then go looking for a dominant sequence and then fill in the issue of the real limit by monotonicity. DCT is often taught only later on, but the concept is simply enough "you are actually allowed to do what you have always wanted to do if you check this condition".
– Phira
Jul 22 at 13:39
In fact the calculus you use here is much more complicated than necessary.
– David C. Ullrich
Jul 22 at 14:03
We can make it more general without restricting $x$ to integers. Starting from your answer, I built mine. By the way, $+1$ for your to compensate a downvote which is not justified (at least for me). Cheers.
– Claude Leibovici
Jul 22 at 14:11
 |Â
show 1 more comment
up vote
-1
down vote
I am not sure if I understood what you wrote. But I am quite sure that you do not need the measure you defined, or in fact any complicated calculus.
The function $f(x):= sumlimits_n=1^infty frac1n(n+x)$ is clearly strictly monotone decreasing and non-negative. So the limit exists, and in fact $limlimits_xrightarrow infty f(x)$ is the limit of the sequence $limlimits_mrightarrow infty f(m)$ where $m$ runs through the positive integers.
So we can restrict our attention to positive integers $x$. Then $frac1n(n+x) = frac1x (frac1n- frac1n+x)$. The sum $sumlimits_n=1^infty frac1n- frac1n+x$ is a telescopic sum, its value is $sumlimits_n=1^x frac1n approx log x$. So $f(x)approx fraclog xx$, which tends to zero as $x$ tends to infinity.
answered Jul 22 at 13:27
A. Pongrácz
2,109221
It's not my question. I don't want to compute this limit. I just want to make a comparaison with measure theory.
– user380364
Jul 22 at 13:28
1
I do not think it can be effectively used to solve this problem. If anything, it seems to complicate matters rather than simplify them.
– A. Pongrácz
Jul 22 at 13:30
On the contrary, it is a very simple approach to see that the terms converge to zero and then go looking for a dominant sequence and then fill in the issue of the real limit by monotonicity. DCT is often taught only later on, but the concept is simply enough "you are actually allowed to do what you have always wanted to do if you check this condition".
– Phira
Jul 22 at 13:39
In fact the calculus you use here is much more complicated than necessary.
– David C. Ullrich
Jul 22 at 14:03
We can make it more general without restricting $x$ to integers. Starting from your answer, I built mine. By the way, $+1$ for your to compensate a downvote which is not justified (at least for me). Cheers.
– Claude Leibovici
Jul 22 at 14:11
 |Â
show 1 more comment
up vote
-1
down vote
up vote
-1
down vote
I am not sure if I understood what you wrote. But I am quite sure that you do not need the measure you defined, or in fact any complicated calculus.
The function $f(x):= sumlimits_n=1^infty frac1n(n+x)$ is clearly strictly monotone decreasing and non-negative. So the limit exists, and in fact $limlimits_xrightarrow infty f(x)$ is the limit of the sequence $limlimits_mrightarrow infty f(m)$ where $m$ runs through the positive integers.
So we can restrict our attention to positive integers $x$. Then $frac1n(n+x) = frac1x (frac1n- frac1n+x)$. The sum $sumlimits_n=1^infty frac1n- frac1n+x$ is a telescopic sum, its value is $sumlimits_n=1^x frac1n approx log x$. So $f(x)approx fraclog xx$, which tends to zero as $x$ tends to infinity.
answered Jul 22 at 13:27
A. Pongrácz
2,109221
I am not sure if I understood what you wrote. But I am quite sure that you do not need the measure you defined, or in fact any complicated calculus.
The function $f(x):= sumlimits_n=1^infty frac1n(n+x)$ is clearly strictly monotone decreasing and non-negative. So the limit exists, and in fact $limlimits_xrightarrow infty f(x)$ is the limit of the sequence $limlimits_mrightarrow infty f(m)$ where $m$ runs through the positive integers.
So we can restrict our attention to positive integers $x$. Then $frac1n(n+x) = frac1x (frac1n- frac1n+x)$. The sum $sumlimits_n=1^infty frac1n- frac1n+x$ is a telescopic sum, its value is $sumlimits_n=1^x frac1n approx log x$. So $f(x)approx fraclog xx$, which tends to zero as $x$ tends to infinity.
answered Jul 22 at 13:27
A. Pongrácz
2,109221
answered Jul 22 at 13:27
A. Pongrácz
2,109221
answered Jul 22 at 13:27
A. Pongrácz
2,109221
answered Jul 22 at 13:27
A. Pongrácz
2,109221
2,109221
It's not my question. I don't want to compute this limit. I just want to make a comparaison with measure theory.
– user380364
Jul 22 at 13:28
1
I do not think it can be effectively used to solve this problem. If anything, it seems to complicate matters rather than simplify them.
– A. Pongrácz
Jul 22 at 13:30
On the contrary, it is a very simple approach to see that the terms converge to zero and then go looking for a dominant sequence and then fill in the issue of the real limit by monotonicity. DCT is often taught only later on, but the concept is simply enough "you are actually allowed to do what you have always wanted to do if you check this condition".
– Phira
Jul 22 at 13:39
In fact the calculus you use here is much more complicated than necessary.
– David C. Ullrich
Jul 22 at 14:03
We can make it more general without restricting $x$ to integers. Starting from your answer, I built mine. By the way, $+1$ for your to compensate a downvote which is not justified (at least for me). Cheers.
– Claude Leibovici
Jul 22 at 14:11
 |Â
show 1 more comment
It's not my question. I don't want to compute this limit. I just want to make a comparaison with measure theory.
– user380364
Jul 22 at 13:28
1
I do not think it can be effectively used to solve this problem. If anything, it seems to complicate matters rather than simplify them.
– A. Pongrácz
Jul 22 at 13:30
On the contrary, it is a very simple approach to see that the terms converge to zero and then go looking for a dominant sequence and then fill in the issue of the real limit by monotonicity. DCT is often taught only later on, but the concept is simply enough "you are actually allowed to do what you have always wanted to do if you check this condition".
– Phira
Jul 22 at 13:39
In fact the calculus you use here is much more complicated than necessary.
– David C. Ullrich
Jul 22 at 14:03
We can make it more general without restricting $x$ to integers. Starting from your answer, I built mine. By the way, $+1$ for your to compensate a downvote which is not justified (at least for me). Cheers.
– Claude Leibovici
Jul 22 at 14:11
It's not my question. I don't want to compute this limit. I just want to make a comparaison with measure theory.
– user380364
Jul 22 at 13:28
It's not my question. I don't want to compute this limit. I just want to make a comparaison with measure theory.
– user380364
Jul 22 at 13:28
1
1
I do not think it can be effectively used to solve this problem. If anything, it seems to complicate matters rather than simplify them.
– A. Pongrácz
Jul 22 at 13:30
I do not think it can be effectively used to solve this problem. If anything, it seems to complicate matters rather than simplify them.
– A. Pongrácz
Jul 22 at 13:30
On the contrary, it is a very simple approach to see that the terms converge to zero and then go looking for a dominant sequence and then fill in the issue of the real limit by monotonicity. DCT is often taught only later on, but the concept is simply enough "you are actually allowed to do what you have always wanted to do if you check this condition".
– Phira
Jul 22 at 13:39
On the contrary, it is a very simple approach to see that the terms converge to zero and then go looking for a dominant sequence and then fill in the issue of the real limit by monotonicity. DCT is often taught only later on, but the concept is simply enough "you are actually allowed to do what you have always wanted to do if you check this condition".
– Phira
Jul 22 at 13:39
In fact the calculus you use here is much more complicated than necessary.
– David C. Ullrich
Jul 22 at 14:03
In fact the calculus you use here is much more complicated than necessary.
– David C. Ullrich
Jul 22 at 14:03
We can make it more general without restricting $x$ to integers. Starting from your answer, I built mine. By the way, $+1$ for your to compensate a downvote which is not justified (at least for me). Cheers.
– Claude Leibovici
Jul 22 at 14:11
We can make it more general without restricting $x$ to integers. Starting from your answer, I built mine. By the way, $+1$ for your to compensate a downvote which is not justified (at least for me). Cheers.
– Claude Leibovici
Jul 22 at 14:11
 |Â
show 1 more comment
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Don’t know if your method works, but I would use M test to prove uniform convergence and then take limit termwise.
– Szeto
Jul 22 at 13:19
1
The dominated convergence theorem is for sequences of functions. In your case you have a limit over the real numbers. You can apply your argument to deduce that for every monotonic sequence $x_mtoinfty$ you have $lim_mtoinftysum_n=1^inftyfrac1n(n+x_m)=0$. Then you need to deduce from that that $lim_xtoinftysum_n=1^inftyfrac1n(n+x)=0$.
– user574889
Jul 22 at 13:23
1
It is important to remember that the dominated convergence theorem is for sequences because of this
– user574889
Jul 22 at 13:27
1
This is a good approach. Use it to show that the limit when $xtoinfty$, $x$ integer, is zero, then extend the result to $xtoinfty$, $x$ real, by monotonicity of the sum with respect to $x$.
– Did
Jul 22 at 13:30
1
@user380364 Monotonicity is not needed. What is important is that on $mathbbR$ a limit $lim_xto a$ exists (as a unique limit) if and only if it exists for all sequences $x_nto a$. This is ultimately due to $mathbbR$ being first countable.
– user574889
Jul 22 at 14:13