Mixing
The limit of $S_n=sum_k=1^n fracknk+n^3$
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Given $n>1$ and $S_n=sumlimits_k=1^n fracknk+n^3$
Calculate $limlimits_n to infty S_n $
Well its obvious that $fracn^2n^3+n leq S_nleq fracn^3n^3+1$
$S_n$ converge to a limit $l$ such that $0leq l leq 1$
How can we determine the value of $l$ ?
sequences-and-series limits
edited Jul 27 at 1:16
Mike Earnest
14.9k11644
asked Jul 27 at 0:57
user515918
284
add a comment |Â
up vote
4
down vote
favorite
Given $n>1$ and $S_n=sumlimits_k=1^n fracknk+n^3$
Calculate $limlimits_n to infty S_n $
Well its obvious that $fracn^2n^3+n leq S_nleq fracn^3n^3+1$
$S_n$ converge to a limit $l$ such that $0leq l leq 1$
How can we determine the value of $l$ ?
sequences-and-series limits
edited Jul 27 at 1:16
Mike Earnest
14.9k11644
asked Jul 27 at 0:57
user515918
284
1
Presumably $k$ is the index of the sum, and not the lower limit, but $n$ is the upper limit? That could be much clearer. A wide hint: See if you can find a way of expressing it as a Riemann Sum.
– Steven Stadnicki
Jul 27 at 1:05
@StevenStadnicki then use the integeration ?
– user515918
Jul 27 at 1:07
Yes, exactly that. Once you've found a way of writing it as a Riemann sum you should be able to turn that into a limit.
– Steven Stadnicki
Jul 27 at 1:08
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given $n>1$ and $S_n=sumlimits_k=1^n fracknk+n^3$
Calculate $limlimits_n to infty S_n $
Well its obvious that $fracn^2n^3+n leq S_nleq fracn^3n^3+1$
$S_n$ converge to a limit $l$ such that $0leq l leq 1$
How can we determine the value of $l$ ?
sequences-and-series limits
edited Jul 27 at 1:16
Mike Earnest
14.9k11644
asked Jul 27 at 0:57
user515918
284
Given $n>1$ and $S_n=sumlimits_k=1^n fracknk+n^3$
Calculate $limlimits_n to infty S_n $
Well its obvious that $fracn^2n^3+n leq S_nleq fracn^3n^3+1$
$S_n$ converge to a limit $l$ such that $0leq l leq 1$
How can we determine the value of $l$ ?
sequences-and-series limits
edited Jul 27 at 1:16
Mike Earnest
14.9k11644
asked Jul 27 at 0:57
user515918
284
edited Jul 27 at 1:16
Mike Earnest
14.9k11644
edited Jul 27 at 1:16
Mike Earnest
14.9k11644
edited Jul 27 at 1:16
Mike Earnest
14.9k11644
14.9k11644
asked Jul 27 at 0:57
user515918
284
asked Jul 27 at 0:57
user515918
284
asked Jul 27 at 0:57
user515918
284
284
1
Presumably $k$ is the index of the sum, and not the lower limit, but $n$ is the upper limit? That could be much clearer. A wide hint: See if you can find a way of expressing it as a Riemann Sum.
– Steven Stadnicki
Jul 27 at 1:05
@StevenStadnicki then use the integeration ?
– user515918
Jul 27 at 1:07
Yes, exactly that. Once you've found a way of writing it as a Riemann sum you should be able to turn that into a limit.
– Steven Stadnicki
Jul 27 at 1:08
add a comment |Â
1
Presumably $k$ is the index of the sum, and not the lower limit, but $n$ is the upper limit? That could be much clearer. A wide hint: See if you can find a way of expressing it as a Riemann Sum.
– Steven Stadnicki
Jul 27 at 1:05
@StevenStadnicki then use the integeration ?
– user515918
Jul 27 at 1:07
Yes, exactly that. Once you've found a way of writing it as a Riemann sum you should be able to turn that into a limit.
– Steven Stadnicki
Jul 27 at 1:08
1
1
Presumably $k$ is the index of the sum, and not the lower limit, but $n$ is the upper limit? That could be much clearer. A wide hint: See if you can find a way of expressing it as a Riemann Sum.
– Steven Stadnicki
Jul 27 at 1:05
Presumably $k$ is the index of the sum, and not the lower limit, but $n$ is the upper limit? That could be much clearer. A wide hint: See if you can find a way of expressing it as a Riemann Sum.
– Steven Stadnicki
Jul 27 at 1:05
@StevenStadnicki then use the integeration ?
– user515918
Jul 27 at 1:07
@StevenStadnicki then use the integeration ?
– user515918
Jul 27 at 1:07
Yes, exactly that. Once you've found a way of writing it as a Riemann sum you should be able to turn that into a limit.
– Steven Stadnicki
Jul 27 at 1:08
Yes, exactly that. Once you've found a way of writing it as a Riemann sum you should be able to turn that into a limit.
– Steven Stadnicki
Jul 27 at 1:08
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
This problem can be tackled by squeeze theorem plus riemann sum.
Note that $$S_n=sum^n_k=1fracfracknfrackn^3+1frac1n$$
We can obtain the following inequality easily
$$sum^n_k=1fracfracknfracnn^3+1frac1nle S_nlesum^n_k=1fracknfrac1n$$
Taking the limit $ntoinfty$, the lower bound approaches
$$lim_ntoinftyleft(frac1frac1n^2+1right)int^1_0xdx=frac12$$
The upper bound approaches $$int^1_0xdx=frac12$$ as well.
By squeeze theorem, $$colorredlim_ntoinftyS_n=frac12$$
(This agrees with your expectation that the limit is between $0$ and $1$.)
answered Jul 27 at 1:18
Szeto
3,9081421
Very very cool!
– Dionel Jaime
Jul 27 at 2:46
@DionelJaime Just looked at your most upvoted post; what a coincidence! So similar to this answer :)
– Szeto
Jul 27 at 3:49
add a comment |Â
up vote
5
down vote
You can also get the result if you only estimate the denominator in order to get better bounds:
$$frac12 stackrelntoinftylongleftarrowfracn(n+1)2(1+n^2) = fracnn+n^3 sum limits_k=1^n k leq S_n leq fracn1+n^3 sum limits_k=1^n k = fracn^2(n+1)2(1+n^3) stackrelntoinftylongrightarrow frac12, . $$
answered Jul 27 at 1:30
ComplexYetTrivial
2,787624
This is a much easier way of doing it; no Riemann sums needed! +1.
– Steven Stadnicki
Jul 27 at 4:45
add a comment |Â
up vote
2
down vote
If you are familiar with harmonic numbers$$S_n=sum_k=1^n fracknk+n^3=n^4 left(H_n^3-H_n^3+nright)+n^2$$ Now, using the asymptotics
$$H_p=gamma +log left(pright)+frac12 p-frac112
p^2+Oleft(frac1p^4right)$$ and applying would give
$$S_n=frac12+frac12 n-frac13 n^2-frac12
n^3+Oleft(frac1n^4right)$$ which shows the limit and how it is approached.
Moreover, this gives a good approximation. For example, the exact value
$$S_10=frac102209264496236445266975187134762733574325443631approx 0.546180$$ while the above expansion would give
$$ frac32776000approx 0.546167$$
answered Jul 27 at 4:34
Claude Leibovici
111k1055126
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
This problem can be tackled by squeeze theorem plus riemann sum.
Note that $$S_n=sum^n_k=1fracfracknfrackn^3+1frac1n$$
We can obtain the following inequality easily
$$sum^n_k=1fracfracknfracnn^3+1frac1nle S_nlesum^n_k=1fracknfrac1n$$
Taking the limit $ntoinfty$, the lower bound approaches
$$lim_ntoinftyleft(frac1frac1n^2+1right)int^1_0xdx=frac12$$
The upper bound approaches $$int^1_0xdx=frac12$$ as well.
By squeeze theorem, $$colorredlim_ntoinftyS_n=frac12$$
(This agrees with your expectation that the limit is between $0$ and $1$.)
answered Jul 27 at 1:18
Szeto
3,9081421
Very very cool!
– Dionel Jaime
Jul 27 at 2:46
@DionelJaime Just looked at your most upvoted post; what a coincidence! So similar to this answer :)
– Szeto
Jul 27 at 3:49
add a comment |Â
up vote
6
down vote
accepted
This problem can be tackled by squeeze theorem plus riemann sum.
Note that $$S_n=sum^n_k=1fracfracknfrackn^3+1frac1n$$
We can obtain the following inequality easily
$$sum^n_k=1fracfracknfracnn^3+1frac1nle S_nlesum^n_k=1fracknfrac1n$$
Taking the limit $ntoinfty$, the lower bound approaches
$$lim_ntoinftyleft(frac1frac1n^2+1right)int^1_0xdx=frac12$$
The upper bound approaches $$int^1_0xdx=frac12$$ as well.
By squeeze theorem, $$colorredlim_ntoinftyS_n=frac12$$
(This agrees with your expectation that the limit is between $0$ and $1$.)
answered Jul 27 at 1:18
Szeto
3,9081421
Very very cool!
– Dionel Jaime
Jul 27 at 2:46
@DionelJaime Just looked at your most upvoted post; what a coincidence! So similar to this answer :)
– Szeto
Jul 27 at 3:49
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
This problem can be tackled by squeeze theorem plus riemann sum.
Note that $$S_n=sum^n_k=1fracfracknfrackn^3+1frac1n$$
We can obtain the following inequality easily
$$sum^n_k=1fracfracknfracnn^3+1frac1nle S_nlesum^n_k=1fracknfrac1n$$
Taking the limit $ntoinfty$, the lower bound approaches
$$lim_ntoinftyleft(frac1frac1n^2+1right)int^1_0xdx=frac12$$
The upper bound approaches $$int^1_0xdx=frac12$$ as well.
By squeeze theorem, $$colorredlim_ntoinftyS_n=frac12$$
(This agrees with your expectation that the limit is between $0$ and $1$.)
answered Jul 27 at 1:18
Szeto
3,9081421
This problem can be tackled by squeeze theorem plus riemann sum.
Note that $$S_n=sum^n_k=1fracfracknfrackn^3+1frac1n$$
We can obtain the following inequality easily
$$sum^n_k=1fracfracknfracnn^3+1frac1nle S_nlesum^n_k=1fracknfrac1n$$
Taking the limit $ntoinfty$, the lower bound approaches
$$lim_ntoinftyleft(frac1frac1n^2+1right)int^1_0xdx=frac12$$
The upper bound approaches $$int^1_0xdx=frac12$$ as well.
By squeeze theorem, $$colorredlim_ntoinftyS_n=frac12$$
(This agrees with your expectation that the limit is between $0$ and $1$.)
answered Jul 27 at 1:18
Szeto
3,9081421
answered Jul 27 at 1:18
Szeto
3,9081421
answered Jul 27 at 1:18
Szeto
3,9081421
answered Jul 27 at 1:18
Szeto
3,9081421
3,9081421
Very very cool!
– Dionel Jaime
Jul 27 at 2:46
@DionelJaime Just looked at your most upvoted post; what a coincidence! So similar to this answer :)
– Szeto
Jul 27 at 3:49
add a comment |Â
Very very cool!
– Dionel Jaime
Jul 27 at 2:46
@DionelJaime Just looked at your most upvoted post; what a coincidence! So similar to this answer :)
– Szeto
Jul 27 at 3:49
Very very cool!
– Dionel Jaime
Jul 27 at 2:46
Very very cool!
– Dionel Jaime
Jul 27 at 2:46
@DionelJaime Just looked at your most upvoted post; what a coincidence! So similar to this answer :)
– Szeto
Jul 27 at 3:49
@DionelJaime Just looked at your most upvoted post; what a coincidence! So similar to this answer :)
– Szeto
Jul 27 at 3:49
add a comment |Â
up vote
5
down vote
You can also get the result if you only estimate the denominator in order to get better bounds:
$$frac12 stackrelntoinftylongleftarrowfracn(n+1)2(1+n^2) = fracnn+n^3 sum limits_k=1^n k leq S_n leq fracn1+n^3 sum limits_k=1^n k = fracn^2(n+1)2(1+n^3) stackrelntoinftylongrightarrow frac12, . $$
answered Jul 27 at 1:30
ComplexYetTrivial
2,787624
This is a much easier way of doing it; no Riemann sums needed! +1.
– Steven Stadnicki
Jul 27 at 4:45
add a comment |Â
up vote
5
down vote
You can also get the result if you only estimate the denominator in order to get better bounds:
$$frac12 stackrelntoinftylongleftarrowfracn(n+1)2(1+n^2) = fracnn+n^3 sum limits_k=1^n k leq S_n leq fracn1+n^3 sum limits_k=1^n k = fracn^2(n+1)2(1+n^3) stackrelntoinftylongrightarrow frac12, . $$
answered Jul 27 at 1:30
ComplexYetTrivial
2,787624
This is a much easier way of doing it; no Riemann sums needed! +1.
– Steven Stadnicki
Jul 27 at 4:45
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You can also get the result if you only estimate the denominator in order to get better bounds:
$$frac12 stackrelntoinftylongleftarrowfracn(n+1)2(1+n^2) = fracnn+n^3 sum limits_k=1^n k leq S_n leq fracn1+n^3 sum limits_k=1^n k = fracn^2(n+1)2(1+n^3) stackrelntoinftylongrightarrow frac12, . $$
answered Jul 27 at 1:30
ComplexYetTrivial
2,787624
You can also get the result if you only estimate the denominator in order to get better bounds:
$$frac12 stackrelntoinftylongleftarrowfracn(n+1)2(1+n^2) = fracnn+n^3 sum limits_k=1^n k leq S_n leq fracn1+n^3 sum limits_k=1^n k = fracn^2(n+1)2(1+n^3) stackrelntoinftylongrightarrow frac12, . $$
answered Jul 27 at 1:30
ComplexYetTrivial
2,787624
answered Jul 27 at 1:30
ComplexYetTrivial
2,787624
answered Jul 27 at 1:30
ComplexYetTrivial
2,787624
answered Jul 27 at 1:30
ComplexYetTrivial
2,787624
2,787624
This is a much easier way of doing it; no Riemann sums needed! +1.
– Steven Stadnicki
Jul 27 at 4:45
add a comment |Â
This is a much easier way of doing it; no Riemann sums needed! +1.
– Steven Stadnicki
Jul 27 at 4:45
This is a much easier way of doing it; no Riemann sums needed! +1.
– Steven Stadnicki
Jul 27 at 4:45
This is a much easier way of doing it; no Riemann sums needed! +1.
– Steven Stadnicki
Jul 27 at 4:45
add a comment |Â
up vote
2
down vote
If you are familiar with harmonic numbers$$S_n=sum_k=1^n fracknk+n^3=n^4 left(H_n^3-H_n^3+nright)+n^2$$ Now, using the asymptotics
$$H_p=gamma +log left(pright)+frac12 p-frac112
p^2+Oleft(frac1p^4right)$$ and applying would give
$$S_n=frac12+frac12 n-frac13 n^2-frac12
n^3+Oleft(frac1n^4right)$$ which shows the limit and how it is approached.
Moreover, this gives a good approximation. For example, the exact value
$$S_10=frac102209264496236445266975187134762733574325443631approx 0.546180$$ while the above expansion would give
$$ frac32776000approx 0.546167$$
answered Jul 27 at 4:34
Claude Leibovici
111k1055126
add a comment |Â
up vote
2
down vote
If you are familiar with harmonic numbers$$S_n=sum_k=1^n fracknk+n^3=n^4 left(H_n^3-H_n^3+nright)+n^2$$ Now, using the asymptotics
$$H_p=gamma +log left(pright)+frac12 p-frac112
p^2+Oleft(frac1p^4right)$$ and applying would give
$$S_n=frac12+frac12 n-frac13 n^2-frac12
n^3+Oleft(frac1n^4right)$$ which shows the limit and how it is approached.
Moreover, this gives a good approximation. For example, the exact value
$$S_10=frac102209264496236445266975187134762733574325443631approx 0.546180$$ while the above expansion would give
$$ frac32776000approx 0.546167$$
answered Jul 27 at 4:34
Claude Leibovici
111k1055126
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you are familiar with harmonic numbers$$S_n=sum_k=1^n fracknk+n^3=n^4 left(H_n^3-H_n^3+nright)+n^2$$ Now, using the asymptotics
$$H_p=gamma +log left(pright)+frac12 p-frac112
p^2+Oleft(frac1p^4right)$$ and applying would give
$$S_n=frac12+frac12 n-frac13 n^2-frac12
n^3+Oleft(frac1n^4right)$$ which shows the limit and how it is approached.
Moreover, this gives a good approximation. For example, the exact value
$$S_10=frac102209264496236445266975187134762733574325443631approx 0.546180$$ while the above expansion would give
$$ frac32776000approx 0.546167$$
answered Jul 27 at 4:34
Claude Leibovici
111k1055126
If you are familiar with harmonic numbers$$S_n=sum_k=1^n fracknk+n^3=n^4 left(H_n^3-H_n^3+nright)+n^2$$ Now, using the asymptotics
$$H_p=gamma +log left(pright)+frac12 p-frac112
p^2+Oleft(frac1p^4right)$$ and applying would give
$$S_n=frac12+frac12 n-frac13 n^2-frac12
n^3+Oleft(frac1n^4right)$$ which shows the limit and how it is approached.
Moreover, this gives a good approximation. For example, the exact value
$$S_10=frac102209264496236445266975187134762733574325443631approx 0.546180$$ while the above expansion would give
$$ frac32776000approx 0.546167$$
answered Jul 27 at 4:34
Claude Leibovici
111k1055126
answered Jul 27 at 4:34
Claude Leibovici
111k1055126
answered Jul 27 at 4:34
Claude Leibovici
111k1055126
answered Jul 27 at 4:34
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863962%2fthe-limit-of-s-n-sum-k-1n-fracknkn3%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Presumably $k$ is the index of the sum, and not the lower limit, but $n$ is the upper limit? That could be much clearer. A wide hint: See if you can find a way of expressing it as a Riemann Sum.
– Steven Stadnicki
Jul 27 at 1:05
@StevenStadnicki then use the integeration ?
– user515918
Jul 27 at 1:07
Yes, exactly that. Once you've found a way of writing it as a Riemann sum you should be able to turn that into a limit.
– Steven Stadnicki
Jul 27 at 1:08