Does the convergence of $$ \sum_{n=0}^{\infty}n a_n $$ imply convergence of:

$$ A = \sum_{n=0}^{\infty} a_n $$

$$ B = \sum_{n=0}^{\infty} \left| a_n\right| $$

$$ C = \sum_{n=0}^{\infty}\sqrt{n} a_n $$

Using the limit criteria it looks like A and C are indeed convergent too. Could you point me in the right direction with sum B?

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    $\begingroup$ On A and C, be careful: the limit comparison test only applies to series of nonnegative terms, so it won't apply here. $\endgroup$ – Daniel Schepler Apr 3 at 15:06
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    $\begingroup$ Hint for what does work on A and C: Dirichlet's test. $\endgroup$ – Daniel Schepler Apr 3 at 15:19
  • $\begingroup$ @DanielSchepler how should one apply Dirichlet's test here? $\endgroup$ – math_beginner Apr 7 at 15:32
  • $\begingroup$ In example C, I have used Dirichlet like this: $\sum_{n=0}^{\infty}\frac{1}{\sqrt{n}} n a_n$. Since $\frac{1}{\sqrt{n}}$ is monotonically decreasing and its limit is 0, and $n a_n$ converges, hence is bounded, the entire series is convergent. After simplifying we end up with expected $\sum_{n=0}^{\infty}\sqrt{n} a_n$. Is that correct assumption? $\endgroup$ – math_beginner Apr 7 at 15:42
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    $\begingroup$ Precisely, $\sum_{n=0}^\infty n a_n$ convergent means exactly the sequence of partial sums $(\sum_{k=0}^n k a_k)_{n=0}^\infty$ is convergent, which implies that the sequence of partial sums is bounded. (Well, technically you need to be careful of the fact that $\frac{1}{\sqrt{n}}$ is undefined at $n=0$ but that's not hard to work around.) $\endgroup$ – Daniel Schepler Apr 7 at 19:12

Try looking at the case $a_n = \frac{(-1)^n}{n \log n}$.

  • $\begingroup$ Great! That's the counter example I was looking for. $\endgroup$ – math_beginner Apr 1 at 16:13

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