I was tasked with proving or disproving the following statement:
If the series $\sum_{n=1}^{\infty} na_{n} $ converges, then $\sum_{n=1}^{\infty} na_{n+1} $ also converges.
I tried to disprove this using $\sum_{n=1}^\infty\frac{1}{n}$ and the fact it diverges, and turning it to $\sum_{n=1}^\infty\frac{1}{n^3} \cdot n$, but that didn't work. Intuitively the statement doesn't sound right because it is too specific.
Write $s_n = \sum_{k=1}^{n} k a_k$. Then we know that $(s_n)$ converges. Now notice that
\begin{align*} \sum_{k=1}^{n} k a_{k+1} &= \sum_{k=1}^{n} \frac{k}{k+1}(s_{k+1} - s_k) \\ &= \sum_{k=1}^{n+1} \frac{k-1}{k} s_k - \sum_{k=1}^{n} \frac{k}{k+1} s_k \\ &= \frac{n}{n+1} s_{n+1} - \sum_{k=1}^{n} \frac{1}{k(k+1)}s_k. \end{align*}
It is straightforward that the last expression converges as $n\to\infty$. As a corollary, we know that $\sum_{n=1}^{\infty} a_n$ converges whenever $\sum_{n=1}^{\infty} n a_n$ converges.