Does $sumlimits_n=1^infty a_nlna_n$ converge?
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up vote
1
down vote
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If $;sumlimits_n=1^infty a_n;$ converges,
does the following also converge? $$sum_n=1^infty a_nlna_n$$
When I first saw this problem, it was easy.
So I tried comparison test and limit comparison test.
But i couldn't solve it.
Could you help me?
sequences-and-series analysis examples-counterexamples
edited Aug 30 at 0:13
Robson
47320
asked May 9 '14 at 14:22
user148928
316213
add a comment |Â
up vote
1
down vote
favorite
If $;sumlimits_n=1^infty a_n;$ converges,
does the following also converge? $$sum_n=1^infty a_nlna_n$$
When I first saw this problem, it was easy.
So I tried comparison test and limit comparison test.
But i couldn't solve it.
Could you help me?
sequences-and-series analysis examples-counterexamples
edited Aug 30 at 0:13
Robson
47320
asked May 9 '14 at 14:22
user148928
316213
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $;sumlimits_n=1^infty a_n;$ converges,
does the following also converge? $$sum_n=1^infty a_nlna_n$$
When I first saw this problem, it was easy.
So I tried comparison test and limit comparison test.
But i couldn't solve it.
Could you help me?
sequences-and-series analysis examples-counterexamples
edited Aug 30 at 0:13
Robson
47320
asked May 9 '14 at 14:22
user148928
316213
If $;sumlimits_n=1^infty a_n;$ converges,
does the following also converge? $$sum_n=1^infty a_nlna_n$$
When I first saw this problem, it was easy.
So I tried comparison test and limit comparison test.
But i couldn't solve it.
Could you help me?
sequences-and-series analysis examples-counterexamples
sequences-and-series analysis examples-counterexamples
edited Aug 30 at 0:13
Robson
47320
asked May 9 '14 at 14:22
user148928
316213
edited Aug 30 at 0:13
Robson
47320
asked May 9 '14 at 14:22
user148928
316213
edited Aug 30 at 0:13
Robson
47320
edited Aug 30 at 0:13
Robson
47320
edited Aug 30 at 0:13
Robson
47320
47320
asked May 9 '14 at 14:22
user148928
316213
asked May 9 '14 at 14:22
user148928
316213
asked May 9 '14 at 14:22
user148928
316213
316213
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
First of all, we have to assume that $a_n>0$ for each $n$, otherwise $log a_n$ wouldn't make sense. The problem is that $a_n$ can go to $0$ fast enough to make the series $sum_n a_n$ convergent, but the fact that $log a_n$ goes to $-infty$ can spoil the convergence.
Indeed, define $a_n:=frac 1n(log n)^2$. Then $sum_n a_n$ is convergent.
On the other hand,
$$|a_nlog a_n|sim frac 1nlog n,$$
hence the series $sum_n a_nlog a_n$ is divergent ($a_nlog a_nleqslant 0$).
answered May 9 '14 at 14:38
Davide Giraudo
122k15147250
How can you find a couterexample easily?
– user148928
May 9 '14 at 14:51
2
We want $sum_n a_n$ to be convergent, but with $a_nto 0$ not too fast because $a_nlog a_n$ will be too small to diverge. Series of the form $sum_n n^-1(log n)^-b$ converge for $b>1$ and diverge otherwise, hence these series are good candidates.
– Davide Giraudo
May 9 '14 at 15:03
Nice! Thank you for your help!
– user148928
May 9 '14 at 15:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First of all, we have to assume that $a_n>0$ for each $n$, otherwise $log a_n$ wouldn't make sense. The problem is that $a_n$ can go to $0$ fast enough to make the series $sum_n a_n$ convergent, but the fact that $log a_n$ goes to $-infty$ can spoil the convergence.
Indeed, define $a_n:=frac 1n(log n)^2$. Then $sum_n a_n$ is convergent.
On the other hand,
$$|a_nlog a_n|sim frac 1nlog n,$$
hence the series $sum_n a_nlog a_n$ is divergent ($a_nlog a_nleqslant 0$).
answered May 9 '14 at 14:38
Davide Giraudo
122k15147250
How can you find a couterexample easily?
– user148928
May 9 '14 at 14:51
2
We want $sum_n a_n$ to be convergent, but with $a_nto 0$ not too fast because $a_nlog a_n$ will be too small to diverge. Series of the form $sum_n n^-1(log n)^-b$ converge for $b>1$ and diverge otherwise, hence these series are good candidates.
– Davide Giraudo
May 9 '14 at 15:03
Nice! Thank you for your help!
– user148928
May 9 '14 at 15:36
add a comment |Â
up vote
2
down vote
accepted
First of all, we have to assume that $a_n>0$ for each $n$, otherwise $log a_n$ wouldn't make sense. The problem is that $a_n$ can go to $0$ fast enough to make the series $sum_n a_n$ convergent, but the fact that $log a_n$ goes to $-infty$ can spoil the convergence.
Indeed, define $a_n:=frac 1n(log n)^2$. Then $sum_n a_n$ is convergent.
On the other hand,
$$|a_nlog a_n|sim frac 1nlog n,$$
hence the series $sum_n a_nlog a_n$ is divergent ($a_nlog a_nleqslant 0$).
answered May 9 '14 at 14:38
Davide Giraudo
122k15147250
How can you find a couterexample easily?
– user148928
May 9 '14 at 14:51
2
We want $sum_n a_n$ to be convergent, but with $a_nto 0$ not too fast because $a_nlog a_n$ will be too small to diverge. Series of the form $sum_n n^-1(log n)^-b$ converge for $b>1$ and diverge otherwise, hence these series are good candidates.
– Davide Giraudo
May 9 '14 at 15:03
Nice! Thank you for your help!
– user148928
May 9 '14 at 15:36
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First of all, we have to assume that $a_n>0$ for each $n$, otherwise $log a_n$ wouldn't make sense. The problem is that $a_n$ can go to $0$ fast enough to make the series $sum_n a_n$ convergent, but the fact that $log a_n$ goes to $-infty$ can spoil the convergence.
Indeed, define $a_n:=frac 1n(log n)^2$. Then $sum_n a_n$ is convergent.
On the other hand,
$$|a_nlog a_n|sim frac 1nlog n,$$
hence the series $sum_n a_nlog a_n$ is divergent ($a_nlog a_nleqslant 0$).
answered May 9 '14 at 14:38
Davide Giraudo
122k15147250
First of all, we have to assume that $a_n>0$ for each $n$, otherwise $log a_n$ wouldn't make sense. The problem is that $a_n$ can go to $0$ fast enough to make the series $sum_n a_n$ convergent, but the fact that $log a_n$ goes to $-infty$ can spoil the convergence.
Indeed, define $a_n:=frac 1n(log n)^2$. Then $sum_n a_n$ is convergent.
On the other hand,
$$|a_nlog a_n|sim frac 1nlog n,$$
hence the series $sum_n a_nlog a_n$ is divergent ($a_nlog a_nleqslant 0$).
answered May 9 '14 at 14:38
Davide Giraudo
122k15147250
answered May 9 '14 at 14:38
Davide Giraudo
122k15147250
answered May 9 '14 at 14:38
Davide Giraudo
122k15147250
answered May 9 '14 at 14:38
Davide Giraudo
122k15147250
122k15147250
How can you find a couterexample easily?
– user148928
May 9 '14 at 14:51
2
We want $sum_n a_n$ to be convergent, but with $a_nto 0$ not too fast because $a_nlog a_n$ will be too small to diverge. Series of the form $sum_n n^-1(log n)^-b$ converge for $b>1$ and diverge otherwise, hence these series are good candidates.
– Davide Giraudo
May 9 '14 at 15:03
Nice! Thank you for your help!
– user148928
May 9 '14 at 15:36
add a comment |Â
How can you find a couterexample easily?
– user148928
May 9 '14 at 14:51
2
We want $sum_n a_n$ to be convergent, but with $a_nto 0$ not too fast because $a_nlog a_n$ will be too small to diverge. Series of the form $sum_n n^-1(log n)^-b$ converge for $b>1$ and diverge otherwise, hence these series are good candidates.
– Davide Giraudo
May 9 '14 at 15:03
Nice! Thank you for your help!
– user148928
May 9 '14 at 15:36
How can you find a couterexample easily?
– user148928
May 9 '14 at 14:51
How can you find a couterexample easily?
– user148928
May 9 '14 at 14:51
2
2
We want $sum_n a_n$ to be convergent, but with $a_nto 0$ not too fast because $a_nlog a_n$ will be too small to diverge. Series of the form $sum_n n^-1(log n)^-b$ converge for $b>1$ and diverge otherwise, hence these series are good candidates.
– Davide Giraudo
May 9 '14 at 15:03
We want $sum_n a_n$ to be convergent, but with $a_nto 0$ not too fast because $a_nlog a_n$ will be too small to diverge. Series of the form $sum_n n^-1(log n)^-b$ converge for $b>1$ and diverge otherwise, hence these series are good candidates.
– Davide Giraudo
May 9 '14 at 15:03
Nice! Thank you for your help!
– user148928
May 9 '14 at 15:36
Nice! Thank you for your help!
– user148928
May 9 '14 at 15:36
add a comment |Â
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