# If the series $\sum_{n=1}^{\infty} na_{n}$ converges, then $\sum_{n=1}^{\infty} na_{n+1}$ also converges? [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 29 '18 at 23:29

• 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. – 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
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.
• @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. – Sangchul Lee Mar 29 '18 at 22:59