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If $sum a_n$ converges and every $a_n$ is positive then $sum a_n^(n-1)/n$ converges?
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Let $(a_n)$ be a sequence of real positive numbers such that $sum a_n$ is a convergent series. What can we say about the series $sum a_n^fracn-1n$? Show that it is convergent or find a counterexample.
I'm trying to find a counterexample but maybe is true that the series is convergent.
sequences-and-series
edited Jul 31 at 13:17
Did
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asked Jul 31 at 10:27
Federico
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up vote
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down vote
favorite
Let $(a_n)$ be a sequence of real positive numbers such that $sum a_n$ is a convergent series. What can we say about the series $sum a_n^fracn-1n$? Show that it is convergent or find a counterexample.
I'm trying to find a counterexample but maybe is true that the series is convergent.
sequences-and-series
edited Jul 31 at 13:17
Did
242k23207441
asked Jul 31 at 10:27
Federico
262
It looks interesting. Write $b_n := a_n^(n-1)/n$. If $$limsup fracb_na_n < +infty,tag1$$ then convergence of $sum b_n$ follows from convergence of $sum a_n$. And the only counterexamples I have found to (1) have $a_n$ going so rapidly to zero that $b_n$ still goes rapidly to zero and $sum b_n$ converges.
– GEdgar
Jul 31 at 11:41
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up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $(a_n)$ be a sequence of real positive numbers such that $sum a_n$ is a convergent series. What can we say about the series $sum a_n^fracn-1n$? Show that it is convergent or find a counterexample.
I'm trying to find a counterexample but maybe is true that the series is convergent.
sequences-and-series
edited Jul 31 at 13:17
Did
242k23207441
asked Jul 31 at 10:27
Federico
262
Let $(a_n)$ be a sequence of real positive numbers such that $sum a_n$ is a convergent series. What can we say about the series $sum a_n^fracn-1n$? Show that it is convergent or find a counterexample.
I'm trying to find a counterexample but maybe is true that the series is convergent.
sequences-and-series
edited Jul 31 at 13:17
Did
242k23207441
asked Jul 31 at 10:27
Federico
262
edited Jul 31 at 13:17
Did
242k23207441
edited Jul 31 at 13:17
Did
242k23207441
edited Jul 31 at 13:17
Did
242k23207441
242k23207441
asked Jul 31 at 10:27
Federico
262
asked Jul 31 at 10:27
Federico
262
asked Jul 31 at 10:27
Federico
262
262
It looks interesting. Write $b_n := a_n^(n-1)/n$. If $$limsup fracb_na_n < +infty,tag1$$ then convergence of $sum b_n$ follows from convergence of $sum a_n$. And the only counterexamples I have found to (1) have $a_n$ going so rapidly to zero that $b_n$ still goes rapidly to zero and $sum b_n$ converges.
– GEdgar
Jul 31 at 11:41
add a comment |Â
It looks interesting. Write $b_n := a_n^(n-1)/n$. If $$limsup fracb_na_n < +infty,tag1$$ then convergence of $sum b_n$ follows from convergence of $sum a_n$. And the only counterexamples I have found to (1) have $a_n$ going so rapidly to zero that $b_n$ still goes rapidly to zero and $sum b_n$ converges.
– GEdgar
Jul 31 at 11:41
It looks interesting. Write $b_n := a_n^(n-1)/n$. If $$limsup fracb_na_n < +infty,tag1$$ then convergence of $sum b_n$ follows from convergence of $sum a_n$. And the only counterexamples I have found to (1) have $a_n$ going so rapidly to zero that $b_n$ still goes rapidly to zero and $sum b_n$ converges.
– GEdgar
Jul 31 at 11:41
It looks interesting. Write $b_n := a_n^(n-1)/n$. If $$limsup fracb_na_n < +infty,tag1$$ then convergence of $sum b_n$ follows from convergence of $sum a_n$. And the only counterexamples I have found to (1) have $a_n$ going so rapidly to zero that $b_n$ still goes rapidly to zero and $sum b_n$ converges.
– GEdgar
Jul 31 at 11:41
add a comment |Â
1 Answer
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$sum_n a_n=sum_n in I a_n +sum_n in J a_n$ where $I=n:a_n <2^-n$ and $J=n:a_n geq 2^-n$. If $n in I$ then $a_n^frac n-1 n<2^-n+1$ so the first sum is convergent. If $n in J$ then $a_n^-1/nleq e^-frac 1 n log frac 1 2^n=2$ so $a_n^frac n-1 n leq 2a_n$ which makes the second series also convergent.
answered Jul 31 at 12:33
Kavi Rama Murthy
19.5k2829
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
$sum_n a_n=sum_n in I a_n +sum_n in J a_n$ where $I=n:a_n <2^-n$ and $J=n:a_n geq 2^-n$. If $n in I$ then $a_n^frac n-1 n<2^-n+1$ so the first sum is convergent. If $n in J$ then $a_n^-1/nleq e^-frac 1 n log frac 1 2^n=2$ so $a_n^frac n-1 n leq 2a_n$ which makes the second series also convergent.
answered Jul 31 at 12:33
Kavi Rama Murthy
19.5k2829
add a comment |Â
up vote
8
down vote
$sum_n a_n=sum_n in I a_n +sum_n in J a_n$ where $I=n:a_n <2^-n$ and $J=n:a_n geq 2^-n$. If $n in I$ then $a_n^frac n-1 n<2^-n+1$ so the first sum is convergent. If $n in J$ then $a_n^-1/nleq e^-frac 1 n log frac 1 2^n=2$ so $a_n^frac n-1 n leq 2a_n$ which makes the second series also convergent.
answered Jul 31 at 12:33
Kavi Rama Murthy
19.5k2829
add a comment |Â
up vote
8
down vote
up vote
8
down vote
$sum_n a_n=sum_n in I a_n +sum_n in J a_n$ where $I=n:a_n <2^-n$ and $J=n:a_n geq 2^-n$. If $n in I$ then $a_n^frac n-1 n<2^-n+1$ so the first sum is convergent. If $n in J$ then $a_n^-1/nleq e^-frac 1 n log frac 1 2^n=2$ so $a_n^frac n-1 n leq 2a_n$ which makes the second series also convergent.
answered Jul 31 at 12:33
Kavi Rama Murthy
19.5k2829
$sum_n a_n=sum_n in I a_n +sum_n in J a_n$ where $I=n:a_n <2^-n$ and $J=n:a_n geq 2^-n$. If $n in I$ then $a_n^frac n-1 n<2^-n+1$ so the first sum is convergent. If $n in J$ then $a_n^-1/nleq e^-frac 1 n log frac 1 2^n=2$ so $a_n^frac n-1 n leq 2a_n$ which makes the second series also convergent.
answered Jul 31 at 12:33
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 12:33
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 12:33
Kavi Rama Murthy
19.5k2829
answered Jul 31 at 12:33
Kavi Rama Murthy
19.5k2829
19.5k2829
add a comment |Â
add a comment |Â
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It looks interesting. Write $b_n := a_n^(n-1)/n$. If $$limsup fracb_na_n < +infty,tag1$$ then convergence of $sum b_n$ follows from convergence of $sum a_n$. And the only counterexamples I have found to (1) have $a_n$ going so rapidly to zero that $b_n$ still goes rapidly to zero and $sum b_n$ converges.
– GEdgar
Jul 31 at 11:41