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.


marked as duplicate by Martin R, A. Goodier, Arnaud D., ncmathsadist, Sangchul Lee calculus Mar 29 '18 at 23:29

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  • $\begingroup$ A partial solution if the $a_n$ are positive (or all negative); by re-indexing we can compare the series $\sum_{n=1}^\infty{na_n}$ and $\sum_{n=2}^\infty{(n-1)a_n}$. Then a limit comparison shows that these have the same convergence/divergence, and the convergence/divergence of the latter series is of course the same as that of $\sum_{n=1}^\infty{na_{n+1}}$. I'm not sure about the case that the $a_n$ are possibly different signs. $\endgroup$ – Hayden Mar 29 '18 at 15:55

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.

  • $\begingroup$ I must confess I dont really understand what you wrote here $\endgroup$ – Bak1139 Mar 29 '18 at 18:32
  • $\begingroup$ @Bak1139, What I did here is to write the partial sum of $\sum n a_{n+1}$ in terms of the partial sum $s_n = \sum_{k=1}^{n}$, using the relation $$ a_{n+1} = \frac{(n+1)a_{n+1}}{n+1} = \frac{s_{n+1} - s_n}{n+1}.$$ Why we are doing this manipulations is because the only information given to us is that $(s_n)$ is convergent. $\endgroup$ – Sangchul Lee Mar 29 '18 at 22:59

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