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Does $sum_k=1^n left(frac1k cdot log(n)right) $ converge to 1?
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I'm trying to show convergence of the following:
$$
limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right)
$$
How shall I proceed?
real-analysis convergence
edited Jul 21 at 1:33
David G. Stork
7,6672929
asked Jul 21 at 0:55
moreblue
1738
add a comment |Â
up vote
0
down vote
favorite
I'm trying to show convergence of the following:
$$
limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right)
$$
How shall I proceed?
real-analysis convergence
edited Jul 21 at 1:33
David G. Stork
7,6672929
asked Jul 21 at 0:55
moreblue
1738
7
You might look up Euler's constant. Also, pull the $log(n)$ out of the sum.
– Ted Shifrin
Jul 21 at 1:00
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to show convergence of the following:
$$
limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right)
$$
How shall I proceed?
real-analysis convergence
edited Jul 21 at 1:33
David G. Stork
7,6672929
asked Jul 21 at 0:55
moreblue
1738
I'm trying to show convergence of the following:
$$
limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right)
$$
How shall I proceed?
real-analysis convergence
edited Jul 21 at 1:33
David G. Stork
7,6672929
asked Jul 21 at 0:55
moreblue
1738
edited Jul 21 at 1:33
David G. Stork
7,6672929
edited Jul 21 at 1:33
David G. Stork
7,6672929
edited Jul 21 at 1:33
David G. Stork
7,6672929
7,6672929
asked Jul 21 at 0:55
moreblue
1738
asked Jul 21 at 0:55
moreblue
1738
asked Jul 21 at 0:55
moreblue
1738
1738
7
You might look up Euler's constant. Also, pull the $log(n)$ out of the sum.
– Ted Shifrin
Jul 21 at 1:00
add a comment |Â
7
You might look up Euler's constant. Also, pull the $log(n)$ out of the sum.
– Ted Shifrin
Jul 21 at 1:00
7
7
You might look up Euler's constant. Also, pull the $log(n)$ out of the sum.
– Ted Shifrin
Jul 21 at 1:00
You might look up Euler's constant. Also, pull the $log(n)$ out of the sum.
– Ted Shifrin
Jul 21 at 1:00
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
Hint: Draw a graph of $f(x) = dfrac1x$ over $[1,n]$,and see that: $dfrac12+dfrac13+cdots +dfrac1n<displaystyle int_1^n dfrac1xdx = ln n < 1+dfrac12+cdots +dfrac1n-1$
answered Jul 21 at 1:51
DeepSea
69k54284
add a comment |Â
up vote
0
down vote
$$frac1log nsum^n_k=1frac1k=fracH_nlog nsimfracgamma+log nlog nto 1$$ as $ntoinfty$.
answered Jul 21 at 2:14
Szeto
4,1631521
add a comment |Â
up vote
0
down vote
Hint
$$S_n=sum_k=1^n frac1k , log(n)=frac1 log(n)sum_k=1^n frac1k=fracH_n log(n)$$ where appears the harmonic number.
Use the asymptotics
$$H_n=gamma +log left(nright)+frac12
n+Oleft(frac1n^2right)$$
answered Jul 21 at 2:16
Claude Leibovici
111k1055126
add a comment |Â
up vote
-1
down vote
Since $log(n)$ can be viewed as a number in the sum:
$$ limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right) = limlimits_n to infty frac1log(n) sum_k=1^n left(frac1k right) = limlimits_n to infty fracsum_k=1^n left(frac1k right)log(n) cdot $$ since $$limlimits_n to infty sum_k=1^n left(frac1k right)$$ diverges so look at this, you can apply L'Hopital Rule (that will be your work).
answered Jul 21 at 2:20
Verónica
235
How do you use L'hopitale rule for this question ? if possible, show it here so OP can see.
– DeepSea
Jul 21 at 2:22
sciweavers.org/upload/Tex2Img_1532140298/render.png I don't know if this works well in this case but it seemed reasonable
– Verónica
Jul 21 at 2:36
Ok, now I tried to figure out this like: lets think this as an integral, if the integral converges then, the sum converges. lets think in this integral: www4d.wolframalpha.com/Calculate/MSP/… it converges to 1, what happens if you do the limit of 1 tending to infinity? Of course! is 1 :)
– Verónica
Jul 21 at 2:59
You can check that here: wolframalpha.com/input/?i=integral++from+1+to+n+1%2F((log(n)*x) and look at that primitive, its very important
– Verónica
Jul 21 at 3:05
I would be grateful if you tell me if you expected that kind of explanation (thinking in integrals) because we could find another if it didn't (like other did in the previous comments with the harmonic function)
– Verónica
Jul 21 at 3:15
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint: Draw a graph of $f(x) = dfrac1x$ over $[1,n]$,and see that: $dfrac12+dfrac13+cdots +dfrac1n<displaystyle int_1^n dfrac1xdx = ln n < 1+dfrac12+cdots +dfrac1n-1$
answered Jul 21 at 1:51
DeepSea
69k54284
add a comment |Â
up vote
2
down vote
Hint: Draw a graph of $f(x) = dfrac1x$ over $[1,n]$,and see that: $dfrac12+dfrac13+cdots +dfrac1n<displaystyle int_1^n dfrac1xdx = ln n < 1+dfrac12+cdots +dfrac1n-1$
answered Jul 21 at 1:51
DeepSea
69k54284
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Draw a graph of $f(x) = dfrac1x$ over $[1,n]$,and see that: $dfrac12+dfrac13+cdots +dfrac1n<displaystyle int_1^n dfrac1xdx = ln n < 1+dfrac12+cdots +dfrac1n-1$
answered Jul 21 at 1:51
DeepSea
69k54284
Hint: Draw a graph of $f(x) = dfrac1x$ over $[1,n]$,and see that: $dfrac12+dfrac13+cdots +dfrac1n<displaystyle int_1^n dfrac1xdx = ln n < 1+dfrac12+cdots +dfrac1n-1$
answered Jul 21 at 1:51
DeepSea
69k54284
answered Jul 21 at 1:51
DeepSea
69k54284
answered Jul 21 at 1:51
DeepSea
69k54284
answered Jul 21 at 1:51
DeepSea
69k54284
69k54284
add a comment |Â
add a comment |Â
up vote
0
down vote
$$frac1log nsum^n_k=1frac1k=fracH_nlog nsimfracgamma+log nlog nto 1$$ as $ntoinfty$.
answered Jul 21 at 2:14
Szeto
4,1631521
add a comment |Â
up vote
0
down vote
$$frac1log nsum^n_k=1frac1k=fracH_nlog nsimfracgamma+log nlog nto 1$$ as $ntoinfty$.
answered Jul 21 at 2:14
Szeto
4,1631521
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$frac1log nsum^n_k=1frac1k=fracH_nlog nsimfracgamma+log nlog nto 1$$ as $ntoinfty$.
answered Jul 21 at 2:14
Szeto
4,1631521
$$frac1log nsum^n_k=1frac1k=fracH_nlog nsimfracgamma+log nlog nto 1$$ as $ntoinfty$.
answered Jul 21 at 2:14
Szeto
4,1631521
answered Jul 21 at 2:14
Szeto
4,1631521
answered Jul 21 at 2:14
Szeto
4,1631521
answered Jul 21 at 2:14
Szeto
4,1631521
4,1631521
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint
$$S_n=sum_k=1^n frac1k , log(n)=frac1 log(n)sum_k=1^n frac1k=fracH_n log(n)$$ where appears the harmonic number.
Use the asymptotics
$$H_n=gamma +log left(nright)+frac12
n+Oleft(frac1n^2right)$$
answered Jul 21 at 2:16
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
Hint
$$S_n=sum_k=1^n frac1k , log(n)=frac1 log(n)sum_k=1^n frac1k=fracH_n log(n)$$ where appears the harmonic number.
Use the asymptotics
$$H_n=gamma +log left(nright)+frac12
n+Oleft(frac1n^2right)$$
answered Jul 21 at 2:16
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint
$$S_n=sum_k=1^n frac1k , log(n)=frac1 log(n)sum_k=1^n frac1k=fracH_n log(n)$$ where appears the harmonic number.
Use the asymptotics
$$H_n=gamma +log left(nright)+frac12
n+Oleft(frac1n^2right)$$
answered Jul 21 at 2:16
Claude Leibovici
111k1055126
Hint
$$S_n=sum_k=1^n frac1k , log(n)=frac1 log(n)sum_k=1^n frac1k=fracH_n log(n)$$ where appears the harmonic number.
Use the asymptotics
$$H_n=gamma +log left(nright)+frac12
n+Oleft(frac1n^2right)$$
answered Jul 21 at 2:16
Claude Leibovici
111k1055126
answered Jul 21 at 2:16
Claude Leibovici
111k1055126
answered Jul 21 at 2:16
Claude Leibovici
111k1055126
answered Jul 21 at 2:16
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
up vote
-1
down vote
Since $log(n)$ can be viewed as a number in the sum:
$$ limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right) = limlimits_n to infty frac1log(n) sum_k=1^n left(frac1k right) = limlimits_n to infty fracsum_k=1^n left(frac1k right)log(n) cdot $$ since $$limlimits_n to infty sum_k=1^n left(frac1k right)$$ diverges so look at this, you can apply L'Hopital Rule (that will be your work).
answered Jul 21 at 2:20
Verónica
235
How do you use L'hopitale rule for this question ? if possible, show it here so OP can see.
– DeepSea
Jul 21 at 2:22
sciweavers.org/upload/Tex2Img_1532140298/render.png I don't know if this works well in this case but it seemed reasonable
– Verónica
Jul 21 at 2:36
Ok, now I tried to figure out this like: lets think this as an integral, if the integral converges then, the sum converges. lets think in this integral: www4d.wolframalpha.com/Calculate/MSP/… it converges to 1, what happens if you do the limit of 1 tending to infinity? Of course! is 1 :)
– Verónica
Jul 21 at 2:59
You can check that here: wolframalpha.com/input/?i=integral++from+1+to+n+1%2F((log(n)*x) and look at that primitive, its very important
– Verónica
Jul 21 at 3:05
I would be grateful if you tell me if you expected that kind of explanation (thinking in integrals) because we could find another if it didn't (like other did in the previous comments with the harmonic function)
– Verónica
Jul 21 at 3:15
add a comment |Â
up vote
-1
down vote
Since $log(n)$ can be viewed as a number in the sum:
$$ limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right) = limlimits_n to infty frac1log(n) sum_k=1^n left(frac1k right) = limlimits_n to infty fracsum_k=1^n left(frac1k right)log(n) cdot $$ since $$limlimits_n to infty sum_k=1^n left(frac1k right)$$ diverges so look at this, you can apply L'Hopital Rule (that will be your work).
answered Jul 21 at 2:20
Verónica
235
How do you use L'hopitale rule for this question ? if possible, show it here so OP can see.
– DeepSea
Jul 21 at 2:22
sciweavers.org/upload/Tex2Img_1532140298/render.png I don't know if this works well in this case but it seemed reasonable
– Verónica
Jul 21 at 2:36
Ok, now I tried to figure out this like: lets think this as an integral, if the integral converges then, the sum converges. lets think in this integral: www4d.wolframalpha.com/Calculate/MSP/… it converges to 1, what happens if you do the limit of 1 tending to infinity? Of course! is 1 :)
– Verónica
Jul 21 at 2:59
You can check that here: wolframalpha.com/input/?i=integral++from+1+to+n+1%2F((log(n)*x) and look at that primitive, its very important
– Verónica
Jul 21 at 3:05
I would be grateful if you tell me if you expected that kind of explanation (thinking in integrals) because we could find another if it didn't (like other did in the previous comments with the harmonic function)
– Verónica
Jul 21 at 3:15
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Since $log(n)$ can be viewed as a number in the sum:
$$ limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right) = limlimits_n to infty frac1log(n) sum_k=1^n left(frac1k right) = limlimits_n to infty fracsum_k=1^n left(frac1k right)log(n) cdot $$ since $$limlimits_n to infty sum_k=1^n left(frac1k right)$$ diverges so look at this, you can apply L'Hopital Rule (that will be your work).
answered Jul 21 at 2:20
Verónica
235
Since $log(n)$ can be viewed as a number in the sum:
$$ limlimits_n to infty sum_k=1^n left(frac1k cdot log(n)right) = limlimits_n to infty frac1log(n) sum_k=1^n left(frac1k right) = limlimits_n to infty fracsum_k=1^n left(frac1k right)log(n) cdot $$ since $$limlimits_n to infty sum_k=1^n left(frac1k right)$$ diverges so look at this, you can apply L'Hopital Rule (that will be your work).
answered Jul 21 at 2:20
Verónica
235
answered Jul 21 at 2:20
Verónica
235
answered Jul 21 at 2:20
Verónica
235
answered Jul 21 at 2:20
Verónica
235
235
How do you use L'hopitale rule for this question ? if possible, show it here so OP can see.
– DeepSea
Jul 21 at 2:22
sciweavers.org/upload/Tex2Img_1532140298/render.png I don't know if this works well in this case but it seemed reasonable
– Verónica
Jul 21 at 2:36
Ok, now I tried to figure out this like: lets think this as an integral, if the integral converges then, the sum converges. lets think in this integral: www4d.wolframalpha.com/Calculate/MSP/… it converges to 1, what happens if you do the limit of 1 tending to infinity? Of course! is 1 :)
– Verónica
Jul 21 at 2:59
You can check that here: wolframalpha.com/input/?i=integral++from+1+to+n+1%2F((log(n)*x) and look at that primitive, its very important
– Verónica
Jul 21 at 3:05
I would be grateful if you tell me if you expected that kind of explanation (thinking in integrals) because we could find another if it didn't (like other did in the previous comments with the harmonic function)
– Verónica
Jul 21 at 3:15
add a comment |Â
How do you use L'hopitale rule for this question ? if possible, show it here so OP can see.
– DeepSea
Jul 21 at 2:22
sciweavers.org/upload/Tex2Img_1532140298/render.png I don't know if this works well in this case but it seemed reasonable
– Verónica
Jul 21 at 2:36
Ok, now I tried to figure out this like: lets think this as an integral, if the integral converges then, the sum converges. lets think in this integral: www4d.wolframalpha.com/Calculate/MSP/… it converges to 1, what happens if you do the limit of 1 tending to infinity? Of course! is 1 :)
– Verónica
Jul 21 at 2:59
You can check that here: wolframalpha.com/input/?i=integral++from+1+to+n+1%2F((log(n)*x) and look at that primitive, its very important
– Verónica
Jul 21 at 3:05
I would be grateful if you tell me if you expected that kind of explanation (thinking in integrals) because we could find another if it didn't (like other did in the previous comments with the harmonic function)
– Verónica
Jul 21 at 3:15
How do you use L'hopitale rule for this question ? if possible, show it here so OP can see.
– DeepSea
Jul 21 at 2:22
How do you use L'hopitale rule for this question ? if possible, show it here so OP can see.
– DeepSea
Jul 21 at 2:22
sciweavers.org/upload/Tex2Img_1532140298/render.png I don't know if this works well in this case but it seemed reasonable
– Verónica
Jul 21 at 2:36
sciweavers.org/upload/Tex2Img_1532140298/render.png I don't know if this works well in this case but it seemed reasonable
– Verónica
Jul 21 at 2:36
Ok, now I tried to figure out this like: lets think this as an integral, if the integral converges then, the sum converges. lets think in this integral: www4d.wolframalpha.com/Calculate/MSP/… it converges to 1, what happens if you do the limit of 1 tending to infinity? Of course! is 1 :)
– Verónica
Jul 21 at 2:59
Ok, now I tried to figure out this like: lets think this as an integral, if the integral converges then, the sum converges. lets think in this integral: www4d.wolframalpha.com/Calculate/MSP/… it converges to 1, what happens if you do the limit of 1 tending to infinity? Of course! is 1 :)
– Verónica
Jul 21 at 2:59
You can check that here: wolframalpha.com/input/?i=integral++from+1+to+n+1%2F((log(n)*x) and look at that primitive, its very important
– Verónica
Jul 21 at 3:05
You can check that here: wolframalpha.com/input/?i=integral++from+1+to+n+1%2F((log(n)*x) and look at that primitive, its very important
– Verónica
Jul 21 at 3:05
I would be grateful if you tell me if you expected that kind of explanation (thinking in integrals) because we could find another if it didn't (like other did in the previous comments with the harmonic function)
– Verónica
Jul 21 at 3:15
I would be grateful if you tell me if you expected that kind of explanation (thinking in integrals) because we could find another if it didn't (like other did in the previous comments with the harmonic function)
– Verónica
Jul 21 at 3:15
add a comment |Â
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7
You might look up Euler's constant. Also, pull the $log(n)$ out of the sum.
– Ted Shifrin
Jul 21 at 1:00