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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.

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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
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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.

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  • $\begingroup$ 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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