Show : $\sum\limits_{n=1}^\infty \frac{a_n}{1+a_n}$ is absolutely convergent [duplicate]

Given $\displaystyle \sum_{n=1}^\infty a_n$ is absolutely convergent. Show that $\displaystyle \sum_{n=1}^\infty \dfrac{a_n}{1+a_n}$ also converges absolutely. (If $a_n \neq -1, \forall n \geq 1$ )

I am not sure where to start with this. Any help would be appreciated

marked as duplicate by Did sequences-and-series 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(); } ); }); }); Apr 24 '16 at 18:09

• You know $a_n\to0$. So there exists $N$ such that $a_n>-1/2$ for all $n>N$... – David C. Ullrich Apr 24 '16 at 17:54

We have: $\dfrac{|a_n|}{|1+a_n|}\leq \dfrac{|a_n|}{1-|a_n|}\leq \dfrac{|a_n|}{1-\frac{1}{2}}= 2|a_n|$ for all $n \geq N_0$ since $a_n \to 0$, and using comparison test, the conclusion follows.
• If $\displaystyle \sum_{k=1}^n |a_k| \to L \implies \displaystyle \sum_{k=1}^n 2|a_k| \to 2L$ – DeepSea Apr 24 '16 at 18:50
From the fact that $\sum_n |a_n|$ converges, one has that $\lim_{n \rightarrow \infty} a_n = 0$. Then you can use the limit comparison test, comparing $\sum_n \bigg|{a_n \over 1 + a_n}\bigg|$ to $\sum_n |a_n|$, to show that your series is absolutely convergent, and therefore convergent.