If $sumlimits_n=1^infty na_n$ converges , does $sumlimits_n=1^infty na_n+1$ converge?
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up vote
14
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I ask for some help with this question:
Prove or provide counter example:
If $sumlimits_n=1^infty na_n$ converges then $sumlimits_n=1^infty na_n+1$ also converges.
I tries this way:
If $sumlimits_n=1^infty na_n$ converges then $na_n to 0$, therefore $a_n to 0$.
There are 3 possible cases:
1) If $a_n >0 $ and $a_n$ is monotonic decreasing sequence then $na_n+1<na_n$ and $sum_n=1^infty na_n+1$ converges by Comparison Test.
2) If $a_n >0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_n+1>a_n$ because in this case $a_n to infty$, therefore it must be $a_n+1 le a_n$ and $sum_n=1^infty na_n+1$ converges by Comparison Test.
3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.
Thanks.
real-analysis sequences-and-series convergence
edited Aug 30 at 0:13
Robson
47320
asked Dec 1 '13 at 12:02
user97484
684414
add a comment |Â
up vote
14
down vote
favorite
I ask for some help with this question:
Prove or provide counter example:
If $sumlimits_n=1^infty na_n$ converges then $sumlimits_n=1^infty na_n+1$ also converges.
I tries this way:
If $sumlimits_n=1^infty na_n$ converges then $na_n to 0$, therefore $a_n to 0$.
There are 3 possible cases:
1) If $a_n >0 $ and $a_n$ is monotonic decreasing sequence then $na_n+1<na_n$ and $sum_n=1^infty na_n+1$ converges by Comparison Test.
2) If $a_n >0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_n+1>a_n$ because in this case $a_n to infty$, therefore it must be $a_n+1 le a_n$ and $sum_n=1^infty na_n+1$ converges by Comparison Test.
3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.
Thanks.
real-analysis sequences-and-series convergence
edited Aug 30 at 0:13
Robson
47320
asked Dec 1 '13 at 12:02
user97484
684414
Yes it does. If $sumna_n$ is convergent, this implies that for all indices greater than $N$, $a_i$ is a decreasing sequence converging to $0$. Hence, for $m>N, a_m>a_m+1$.
– fierydemon
Dec 1 '13 at 12:06
@AyushKhaitan, I doubt that that is true. $a_i$'s could slightly oscillate.
– Listing
Dec 1 '13 at 12:09
My statement still stands.
– fierydemon
Dec 1 '13 at 12:11
5
If you rewrite the second sum in the form $sum_n=2^infty (n-1)a_n$, you see that it is equivalent to determine whether or not $sum a_n$ must converge. This seems a bit easier to tackle. In particular, if $a_nge0$ then the answer is obviouslyl yes, by the comparison principle.
– Harald Hanche-Olsen
Dec 1 '13 at 12:16
@Macavity If you use such an $a_n$, $sum_n=1^infty na_n$ won't converge.
– user10444
Dec 1 '13 at 12:26
add a comment |Â
up vote
14
down vote
favorite
up vote
14
down vote
favorite
I ask for some help with this question:
Prove or provide counter example:
If $sumlimits_n=1^infty na_n$ converges then $sumlimits_n=1^infty na_n+1$ also converges.
I tries this way:
If $sumlimits_n=1^infty na_n$ converges then $na_n to 0$, therefore $a_n to 0$.
There are 3 possible cases:
1) If $a_n >0 $ and $a_n$ is monotonic decreasing sequence then $na_n+1<na_n$ and $sum_n=1^infty na_n+1$ converges by Comparison Test.
2) If $a_n >0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_n+1>a_n$ because in this case $a_n to infty$, therefore it must be $a_n+1 le a_n$ and $sum_n=1^infty na_n+1$ converges by Comparison Test.
3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.
Thanks.
real-analysis sequences-and-series convergence
edited Aug 30 at 0:13
Robson
47320
asked Dec 1 '13 at 12:02
user97484
684414
I ask for some help with this question:
Prove or provide counter example:
If $sumlimits_n=1^infty na_n$ converges then $sumlimits_n=1^infty na_n+1$ also converges.
I tries this way:
If $sumlimits_n=1^infty na_n$ converges then $na_n to 0$, therefore $a_n to 0$.
There are 3 possible cases:
1) If $a_n >0 $ and $a_n$ is monotonic decreasing sequence then $na_n+1<na_n$ and $sum_n=1^infty na_n+1$ converges by Comparison Test.
2) If $a_n >0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_n+1>a_n$ because in this case $a_n to infty$, therefore it must be $a_n+1 le a_n$ and $sum_n=1^infty na_n+1$ converges by Comparison Test.
3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.
Thanks.
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Aug 30 at 0:13
Robson
47320
asked Dec 1 '13 at 12:02
user97484
684414
edited Aug 30 at 0:13
Robson
47320
asked Dec 1 '13 at 12:02
user97484
684414
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 Dec 1 '13 at 12:02
user97484
684414
asked Dec 1 '13 at 12:02
user97484
684414
asked Dec 1 '13 at 12:02
user97484
684414
684414
Yes it does. If $sumna_n$ is convergent, this implies that for all indices greater than $N$, $a_i$ is a decreasing sequence converging to $0$. Hence, for $m>N, a_m>a_m+1$.
– fierydemon
Dec 1 '13 at 12:06
@AyushKhaitan, I doubt that that is true. $a_i$'s could slightly oscillate.
– Listing
Dec 1 '13 at 12:09
My statement still stands.
– fierydemon
Dec 1 '13 at 12:11
5
If you rewrite the second sum in the form $sum_n=2^infty (n-1)a_n$, you see that it is equivalent to determine whether or not $sum a_n$ must converge. This seems a bit easier to tackle. In particular, if $a_nge0$ then the answer is obviouslyl yes, by the comparison principle.
– Harald Hanche-Olsen
Dec 1 '13 at 12:16
@Macavity If you use such an $a_n$, $sum_n=1^infty na_n$ won't converge.
– user10444
Dec 1 '13 at 12:26
add a comment |Â
Yes it does. If $sumna_n$ is convergent, this implies that for all indices greater than $N$, $a_i$ is a decreasing sequence converging to $0$. Hence, for $m>N, a_m>a_m+1$.
– fierydemon
Dec 1 '13 at 12:06
@AyushKhaitan, I doubt that that is true. $a_i$'s could slightly oscillate.
– Listing
Dec 1 '13 at 12:09
My statement still stands.
– fierydemon
Dec 1 '13 at 12:11
5
If you rewrite the second sum in the form $sum_n=2^infty (n-1)a_n$, you see that it is equivalent to determine whether or not $sum a_n$ must converge. This seems a bit easier to tackle. In particular, if $a_nge0$ then the answer is obviouslyl yes, by the comparison principle.
– Harald Hanche-Olsen
Dec 1 '13 at 12:16
@Macavity If you use such an $a_n$, $sum_n=1^infty na_n$ won't converge.
– user10444
Dec 1 '13 at 12:26
Yes it does. If $sumna_n$ is convergent, this implies that for all indices greater than $N$, $a_i$ is a decreasing sequence converging to $0$. Hence, for $m>N, a_m>a_m+1$.
– fierydemon
Dec 1 '13 at 12:06
Yes it does. If $sumna_n$ is convergent, this implies that for all indices greater than $N$, $a_i$ is a decreasing sequence converging to $0$. Hence, for $m>N, a_m>a_m+1$.
– fierydemon
Dec 1 '13 at 12:06
@AyushKhaitan, I doubt that that is true. $a_i$'s could slightly oscillate.
– Listing
Dec 1 '13 at 12:09
@AyushKhaitan, I doubt that that is true. $a_i$'s could slightly oscillate.
– Listing
Dec 1 '13 at 12:09
My statement still stands.
– fierydemon
Dec 1 '13 at 12:11
My statement still stands.
– fierydemon
Dec 1 '13 at 12:11
5
5
If you rewrite the second sum in the form $sum_n=2^infty (n-1)a_n$, you see that it is equivalent to determine whether or not $sum a_n$ must converge. This seems a bit easier to tackle. In particular, if $a_nge0$ then the answer is obviouslyl yes, by the comparison principle.
– Harald Hanche-Olsen
Dec 1 '13 at 12:16
If you rewrite the second sum in the form $sum_n=2^infty (n-1)a_n$, you see that it is equivalent to determine whether or not $sum a_n$ must converge. This seems a bit easier to tackle. In particular, if $a_nge0$ then the answer is obviouslyl yes, by the comparison principle.
– Harald Hanche-Olsen
Dec 1 '13 at 12:16
@Macavity If you use such an $a_n$, $sum_n=1^infty na_n$ won't converge.
– user10444
Dec 1 '13 at 12:26
@Macavity If you use such an $a_n$, $sum_n=1^infty na_n$ won't converge.
– user10444
Dec 1 '13 at 12:26
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
16
down vote
accepted
Yes. Put $b_n=na_n$, so the question is now (see my comment on the question):
If $displaystylesum_n=1^infty b_n$ converges, does $displaystylesum_n=1^inftyfracb_nn$ converge?
Let $s_n=sum_k=1^n b_n$. We get (partial summation)
$$
sum_k=1^nfracb_kk
=sum_k=1^nfracs_k-s_k-1k
=sum_k=1^nBigl(frac1k-frac1k+1Bigr)s_k+fracs_nn+1
=sum_k=1^nfrac1k(k+1)s_k+fracs_nn+1
$$
which converges as $ntoinfty$, because $s_k$ is bounded, so the sum is absolutely convergent.
answered Dec 1 '13 at 12:37
Harald Hanche-Olsen
27.3k23959
7
I do feel a little bad about posting such a complete answer to a homework question, yet it seemed that an antidote was needed to all the bad answers out there.
– Harald Hanche-Olsen
Dec 1 '13 at 12:40
add a comment |Â
up vote
5
down vote
Using the abelian and tauberian theorem seems reasonable. Let
$$f(z)=sum a_nz^n$$
Now, if
$$sum n a_n$$
converges, then
$$f'(z) to sum n a_n$$
when $z to 1^-$ (abelian theorem). Then
$$int_0^zf'(u)du$$
tends to a definite value when $z to 1^-.$ Recall that $a_n=o(1/n)$. The tauberian theorem then asserts that
$$f(z) to sum a_n$$
when $z to 1^-$ in such a case, and hence we are done.
edited Aug 30 at 0:27
Mattos
2,66721121
answered Dec 1 '13 at 12:53
y zh
1,8971321
2
That is quite clever.
– Harald Hanche-Olsen
Dec 1 '13 at 13:02
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
16
down vote
accepted
Yes. Put $b_n=na_n$, so the question is now (see my comment on the question):
If $displaystylesum_n=1^infty b_n$ converges, does $displaystylesum_n=1^inftyfracb_nn$ converge?
Let $s_n=sum_k=1^n b_n$. We get (partial summation)
$$
sum_k=1^nfracb_kk
=sum_k=1^nfracs_k-s_k-1k
=sum_k=1^nBigl(frac1k-frac1k+1Bigr)s_k+fracs_nn+1
=sum_k=1^nfrac1k(k+1)s_k+fracs_nn+1
$$
which converges as $ntoinfty$, because $s_k$ is bounded, so the sum is absolutely convergent.
answered Dec 1 '13 at 12:37
Harald Hanche-Olsen
27.3k23959
7
I do feel a little bad about posting such a complete answer to a homework question, yet it seemed that an antidote was needed to all the bad answers out there.
– Harald Hanche-Olsen
Dec 1 '13 at 12:40
add a comment |Â
up vote
16
down vote
accepted
Yes. Put $b_n=na_n$, so the question is now (see my comment on the question):
If $displaystylesum_n=1^infty b_n$ converges, does $displaystylesum_n=1^inftyfracb_nn$ converge?
Let $s_n=sum_k=1^n b_n$. We get (partial summation)
$$
sum_k=1^nfracb_kk
=sum_k=1^nfracs_k-s_k-1k
=sum_k=1^nBigl(frac1k-frac1k+1Bigr)s_k+fracs_nn+1
=sum_k=1^nfrac1k(k+1)s_k+fracs_nn+1
$$
which converges as $ntoinfty$, because $s_k$ is bounded, so the sum is absolutely convergent.
answered Dec 1 '13 at 12:37
Harald Hanche-Olsen
27.3k23959
7
I do feel a little bad about posting such a complete answer to a homework question, yet it seemed that an antidote was needed to all the bad answers out there.
– Harald Hanche-Olsen
Dec 1 '13 at 12:40
add a comment |Â
up vote
16
down vote
accepted
up vote
16
down vote
accepted
Yes. Put $b_n=na_n$, so the question is now (see my comment on the question):
If $displaystylesum_n=1^infty b_n$ converges, does $displaystylesum_n=1^inftyfracb_nn$ converge?
Let $s_n=sum_k=1^n b_n$. We get (partial summation)
$$
sum_k=1^nfracb_kk
=sum_k=1^nfracs_k-s_k-1k
=sum_k=1^nBigl(frac1k-frac1k+1Bigr)s_k+fracs_nn+1
=sum_k=1^nfrac1k(k+1)s_k+fracs_nn+1
$$
which converges as $ntoinfty$, because $s_k$ is bounded, so the sum is absolutely convergent.
answered Dec 1 '13 at 12:37
Harald Hanche-Olsen
27.3k23959
Yes. Put $b_n=na_n$, so the question is now (see my comment on the question):
If $displaystylesum_n=1^infty b_n$ converges, does $displaystylesum_n=1^inftyfracb_nn$ converge?
Let $s_n=sum_k=1^n b_n$. We get (partial summation)
$$
sum_k=1^nfracb_kk
=sum_k=1^nfracs_k-s_k-1k
=sum_k=1^nBigl(frac1k-frac1k+1Bigr)s_k+fracs_nn+1
=sum_k=1^nfrac1k(k+1)s_k+fracs_nn+1
$$
which converges as $ntoinfty$, because $s_k$ is bounded, so the sum is absolutely convergent.
answered Dec 1 '13 at 12:37
Harald Hanche-Olsen
27.3k23959
answered Dec 1 '13 at 12:37
Harald Hanche-Olsen
27.3k23959
answered Dec 1 '13 at 12:37
Harald Hanche-Olsen
27.3k23959
answered Dec 1 '13 at 12:37
Harald Hanche-Olsen
27.3k23959
27.3k23959
7
I do feel a little bad about posting such a complete answer to a homework question, yet it seemed that an antidote was needed to all the bad answers out there.
– Harald Hanche-Olsen
Dec 1 '13 at 12:40
add a comment |Â
7
I do feel a little bad about posting such a complete answer to a homework question, yet it seemed that an antidote was needed to all the bad answers out there.
– Harald Hanche-Olsen
Dec 1 '13 at 12:40
7
7
I do feel a little bad about posting such a complete answer to a homework question, yet it seemed that an antidote was needed to all the bad answers out there.
– Harald Hanche-Olsen
Dec 1 '13 at 12:40
I do feel a little bad about posting such a complete answer to a homework question, yet it seemed that an antidote was needed to all the bad answers out there.
– Harald Hanche-Olsen
Dec 1 '13 at 12:40
add a comment |Â
up vote
5
down vote
Using the abelian and tauberian theorem seems reasonable. Let
$$f(z)=sum a_nz^n$$
Now, if
$$sum n a_n$$
converges, then
$$f'(z) to sum n a_n$$
when $z to 1^-$ (abelian theorem). Then
$$int_0^zf'(u)du$$
tends to a definite value when $z to 1^-.$ Recall that $a_n=o(1/n)$. The tauberian theorem then asserts that
$$f(z) to sum a_n$$
when $z to 1^-$ in such a case, and hence we are done.
edited Aug 30 at 0:27
Mattos
2,66721121
answered Dec 1 '13 at 12:53
y zh
1,8971321
2
That is quite clever.
– Harald Hanche-Olsen
Dec 1 '13 at 13:02
add a comment |Â
up vote
5
down vote
Using the abelian and tauberian theorem seems reasonable. Let
$$f(z)=sum a_nz^n$$
Now, if
$$sum n a_n$$
converges, then
$$f'(z) to sum n a_n$$
when $z to 1^-$ (abelian theorem). Then
$$int_0^zf'(u)du$$
tends to a definite value when $z to 1^-.$ Recall that $a_n=o(1/n)$. The tauberian theorem then asserts that
$$f(z) to sum a_n$$
when $z to 1^-$ in such a case, and hence we are done.
edited Aug 30 at 0:27
Mattos
2,66721121
answered Dec 1 '13 at 12:53
y zh
1,8971321
2
That is quite clever.
– Harald Hanche-Olsen
Dec 1 '13 at 13:02
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Using the abelian and tauberian theorem seems reasonable. Let
$$f(z)=sum a_nz^n$$
Now, if
$$sum n a_n$$
converges, then
$$f'(z) to sum n a_n$$
when $z to 1^-$ (abelian theorem). Then
$$int_0^zf'(u)du$$
tends to a definite value when $z to 1^-.$ Recall that $a_n=o(1/n)$. The tauberian theorem then asserts that
$$f(z) to sum a_n$$
when $z to 1^-$ in such a case, and hence we are done.
edited Aug 30 at 0:27
Mattos
2,66721121
answered Dec 1 '13 at 12:53
y zh
1,8971321
Using the abelian and tauberian theorem seems reasonable. Let
$$f(z)=sum a_nz^n$$
Now, if
$$sum n a_n$$
converges, then
$$f'(z) to sum n a_n$$
when $z to 1^-$ (abelian theorem). Then
$$int_0^zf'(u)du$$
tends to a definite value when $z to 1^-.$ Recall that $a_n=o(1/n)$. The tauberian theorem then asserts that
$$f(z) to sum a_n$$
when $z to 1^-$ in such a case, and hence we are done.
edited Aug 30 at 0:27
Mattos
2,66721121
answered Dec 1 '13 at 12:53
y zh
1,8971321
edited Aug 30 at 0:27
Mattos
2,66721121
edited Aug 30 at 0:27
Mattos
2,66721121
edited Aug 30 at 0:27
Mattos
2,66721121
2,66721121
answered Dec 1 '13 at 12:53
y zh
1,8971321
answered Dec 1 '13 at 12:53
y zh
1,8971321
answered Dec 1 '13 at 12:53
y zh
1,8971321
1,8971321
2
That is quite clever.
– Harald Hanche-Olsen
Dec 1 '13 at 13:02
add a comment |Â
2
That is quite clever.
– Harald Hanche-Olsen
Dec 1 '13 at 13:02
2
2
That is quite clever.
– Harald Hanche-Olsen
Dec 1 '13 at 13:02
That is quite clever.
– Harald Hanche-Olsen
Dec 1 '13 at 13:02
add a comment |Â
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Yes it does. If $sumna_n$ is convergent, this implies that for all indices greater than $N$, $a_i$ is a decreasing sequence converging to $0$. Hence, for $m>N, a_m>a_m+1$.
– fierydemon
Dec 1 '13 at 12:06
@AyushKhaitan, I doubt that that is true. $a_i$'s could slightly oscillate.
– Listing
Dec 1 '13 at 12:09
My statement still stands.
– fierydemon
Dec 1 '13 at 12:11
5
If you rewrite the second sum in the form $sum_n=2^infty (n-1)a_n$, you see that it is equivalent to determine whether or not $sum a_n$ must converge. This seems a bit easier to tackle. In particular, if $a_nge0$ then the answer is obviouslyl yes, by the comparison principle.
– Harald Hanche-Olsen
Dec 1 '13 at 12:16
@Macavity If you use such an $a_n$, $sum_n=1^infty na_n$ won't converge.
– user10444
Dec 1 '13 at 12:26