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Convergence of a series involving logs $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$
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I am working on showing the convergence of the series $sum_{n=2}^{infty} frac{sin(nx)}{ln(n)}$ and have reduced the problem to showing that $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$ converges. I have a hunch that it does converge, but am having trouble manipulating it to something I can apply the comparison test to. Hints to get me started are appreciated.
sequences-and-series
edited Jan 14 at 21:58
Bernard
121k740116
asked Jan 14 at 20:25
JungleshrimpJungleshrimp
314111
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add a comment |
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I am working on showing the convergence of the series $sum_{n=2}^{infty} frac{sin(nx)}{ln(n)}$ and have reduced the problem to showing that $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$ converges. I have a hunch that it does converge, but am having trouble manipulating it to something I can apply the comparison test to. Hints to get me started are appreciated.
sequences-and-series
edited Jan 14 at 21:58
Bernard
121k740116
asked Jan 14 at 20:25
JungleshrimpJungleshrimp
314111
$endgroup$
4
$begingroup$
You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
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– p4sch
Jan 14 at 20:27
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Could you write down how did you reduce the series from first form to the second one?
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– user
Jan 14 at 22:07
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In the denominator, is it the log squared or the log of the square?
$endgroup$
– Bernard
Jan 14 at 22:28
add a comment |
$begingroup$
I am working on showing the convergence of the series $sum_{n=2}^{infty} frac{sin(nx)}{ln(n)}$ and have reduced the problem to showing that $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$ converges. I have a hunch that it does converge, but am having trouble manipulating it to something I can apply the comparison test to. Hints to get me started are appreciated.
sequences-and-series
edited Jan 14 at 21:58
Bernard
121k740116
asked Jan 14 at 20:25
JungleshrimpJungleshrimp
314111
$endgroup$
I am working on showing the convergence of the series $sum_{n=2}^{infty} frac{sin(nx)}{ln(n)}$ and have reduced the problem to showing that $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$ converges. I have a hunch that it does converge, but am having trouble manipulating it to something I can apply the comparison test to. Hints to get me started are appreciated.
sequences-and-series
sequences-and-series
edited Jan 14 at 21:58
Bernard
121k740116
asked Jan 14 at 20:25
JungleshrimpJungleshrimp
314111
edited Jan 14 at 21:58
Bernard
121k740116
asked Jan 14 at 20:25
JungleshrimpJungleshrimp
314111
edited Jan 14 at 21:58
Bernard
121k740116
edited Jan 14 at 21:58
Bernard
121k740116
edited Jan 14 at 21:58
Bernard
121k740116
121k740116
asked Jan 14 at 20:25
JungleshrimpJungleshrimp
314111
asked Jan 14 at 20:25
JungleshrimpJungleshrimp
314111
asked Jan 14 at 20:25
JungleshrimpJungleshrimp
314111
314111
4
$begingroup$
You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
$endgroup$
– p4sch
Jan 14 at 20:27
$begingroup$
Could you write down how did you reduce the series from first form to the second one?
$endgroup$
– user
Jan 14 at 22:07
$begingroup$
In the denominator, is it the log squared or the log of the square?
$endgroup$
– Bernard
Jan 14 at 22:28
add a comment |
4
$begingroup$
You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
$endgroup$
– p4sch
Jan 14 at 20:27
$begingroup$
Could you write down how did you reduce the series from first form to the second one?
$endgroup$
– user
Jan 14 at 22:07
$begingroup$
In the denominator, is it the log squared or the log of the square?
$endgroup$
– Bernard
Jan 14 at 22:28
4
4
$begingroup$
You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
$endgroup$
– p4sch
Jan 14 at 20:27
$begingroup$
You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
$endgroup$
– p4sch
Jan 14 at 20:27
$begingroup$
Could you write down how did you reduce the series from first form to the second one?
$endgroup$
– user
Jan 14 at 22:07
$begingroup$
Could you write down how did you reduce the series from first form to the second one?
$endgroup$
– user
Jan 14 at 22:07
$begingroup$
In the denominator, is it the log squared or the log of the square?
$endgroup$
– Bernard
Jan 14 at 22:28
$begingroup$
In the denominator, is it the log squared or the log of the square?
$endgroup$
– Bernard
Jan 14 at 22:28
add a comment |
1 Answer
1
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Supposing the denominator is the log squared, we can use equivalence:
$$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
which is Bertrand's series.
This Bertrand's series converges by the integral test.
answered Jan 14 at 22:46
BernardBernard
121k740116
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Supposing the denominator is the log squared, we can use equivalence:
$$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
which is Bertrand's series.
This Bertrand's series converges by the integral test.
answered Jan 14 at 22:46
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Supposing the denominator is the log squared, we can use equivalence:
$$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
which is Bertrand's series.
This Bertrand's series converges by the integral test.
answered Jan 14 at 22:46
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Supposing the denominator is the log squared, we can use equivalence:
$$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
which is Bertrand's series.
This Bertrand's series converges by the integral test.
answered Jan 14 at 22:46
BernardBernard
121k740116
$endgroup$
Supposing the denominator is the log squared, we can use equivalence:
$$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
which is Bertrand's series.
This Bertrand's series converges by the integral test.
answered Jan 14 at 22:46
BernardBernard
121k740116
answered Jan 14 at 22:46
BernardBernard
121k740116
answered Jan 14 at 22:46
BernardBernard
121k740116
answered Jan 14 at 22:46
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
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4
$begingroup$
You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
$endgroup$
– p4sch
Jan 14 at 20:27
$begingroup$
Could you write down how did you reduce the series from first form to the second one?
$endgroup$
– user
Jan 14 at 22:07
$begingroup$
In the denominator, is it the log squared or the log of the square?
$endgroup$
– Bernard
Jan 14 at 22:28