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

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?

• On A and C, be careful: the limit comparison test only applies to series of nonnegative terms, so it won't apply here. – Daniel Schepler Apr 3 at 15:06
• Hint for what does work on A and C: Dirichlet's test. – Daniel Schepler Apr 3 at 15:19
• @DanielSchepler how should one apply Dirichlet's test here? – math_beginner Apr 7 at 15:32
• 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? – math_beginner Apr 7 at 15:42
• 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.) – Daniel Schepler Apr 7 at 19:12

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