Does $\sum na_n$ converge implies $\sum a_n$ converges

Asked Sep 19 '21 at 15:23

Active Sep 19 '21 at 15:46

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If $(a_n)$ is a complex sequence, and $\sum na_n$ converges, does $\sum a_n$ converge?

I found myself a solution but it seems to me a little bit complicated, and I am not sure that I am right.

Let $f(x) = \sum_{n>0} na_n x^{n-1}$.

$f(1)$ is defined so by Abel's Theorem, $f$ is ~~normally~~ uniformally convergent and continuous on $[0,1]$ and therefore

$$ \int_0^1 \sum na_n x^{n-1}dx = \sum \int_0^1 na_nx^{n-1}dx = \sum a_n $$ which concludes

Is there a easier or more direct way to conclude?

Thank you in advance.

edited Sep 19 '21 at 15:46

asked Sep 19 '21 at 15:23

lioneltour

2

Hint: use the fact that if $n \geqslant 1$, then $|a_n|\leqslant n |a_n| =|na_n|$. (your solution is quite an overkill one!) – Didier Sep 19 '21 at 15:27
Here is another solution: https://math.stackexchange.com/a/3024889/42969 – Martin R Sep 19 '21 at 15:38
Or this one: https://math.stackexchange.com/q/603297/42969 (it talks about real series, but Dirichlet's test works for complex series as well). – Martin R Sep 19 '21 at 15:39
1

Thank you Martin ;) – lioneltour Sep 19 '21 at 16:00

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