How to prove that $\sum_{n=1}^{\infty}\frac{a_{n}}{1+a_{n}}$ converges absolutely If $\sum_{n=1}^{\infty}a_{n}$ converges absolutely, show that
$$\sum_{n=1}^{\infty}\dfrac{a_{n}}{1+a_{n}}$$ converges absolutely.
My try: since $\sum_{n=1}^{\infty}a_{n}$ converges absolutely, then
$$\sum_{n=1}^{\infty}|a_{n}|$$ converges, then there exsit $M>0$, such
$$\sum_{n=1}^{N}|a_{n}|<M$$
then, how to prove than
$$\sum_{n=1}^{N}\dfrac{|a_{n}|}{|1+a_{n}|}<cM?$$
where $c$ is a constant?
 A: Hint: Prove that if $\sum a_n$ converges absolutely and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges absolutely.
A: In the case that $a_n$ is complex, we notice that
$$\lim_{n\rightarrow\infty}|a_n|=0,$$
so there is $N>0$ such that
$$|a_n|<\frac{1}{2},\quad n\geq N.$$
Then $|1+a_n|\geq 1-|a_n|>\frac{1}{2}$ for $n\geq N$. Hence
$$\sum_{n=0}^{\infty}\frac{|a_n|}{|1+a_n|}=\sum_{n=0}^{N}\frac{|a_n|}{|1+a_n|}+\sum_{n=N+1}^{\infty}\frac{|a_n|}{|1+a_n|}
\leq\sum_{n=0}^{N}\frac{|a_n|}{|1+a_n|}+2\sum_{n=N+1}^{\infty}|a_n|<+\infty.$$
Hence the series converges absolutly.
A: Of course none of this makes sense if $a_n = -1$ for some $n$, so we'll just assume that it doesn't.
First note that $\displaystyle \lim_{n\rightarrow\infty} a_n = 0$.
Now apply the limit comparison test to the series $\displaystyle \sum_{n=0}^\infty |a_n|$ and $\displaystyle \sum_{n=0}^\infty b_n$ where $\displaystyle b_n = \frac{|a_n|}{|1+a_n|}$.  Then $\displaystyle \lim_{n\rightarrow\infty} \frac{a_n}{b_n} = \lim_{n\rightarrow\infty}|1+a_n| = 1$ since $\displaystyle \lim_{n\rightarrow 0} a_n = 0$, so $\displaystyle \sum_{n=0}^\infty \frac{|a_n|}{|1+a_n|}$ converges by the limit comparison test.
A: You have trouble if $a_n$ is ever equal to $-1$.
Otherwise, since the first series converges, $a_n\to 0$ as $n\to\infty$.
So $a_n$ is eventually greater than $-1/2$.  Say, for all $n>N$.
So $\left|\frac{a_n}{1+a_n}\right|<2|a_n|$ after that point.
Can you put $n\leq N$ and $n>N$ together to show the sum converges?
A: It is well-known that 
$$
\frac{x}{x+1}<x
$$
for positive x. Therefore we have for each term that $|a_n|\geq \frac{|a_n|}{|1+a_n|}$. Therefore 
$$
\infty >\sum_{n=0}^\infty |a_n| \geq \sum_{n=0}^\infty \frac{|a_n|}{|1+a_n|}
$$
And the sum converges absolutely.
