# 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?

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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.

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Hint: Prove that if $\sum a_n$ converges absolutely and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges absolutely.

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For reference, this is known as Abel's test –  Omnomnomnom Dec 9 '13 at 13:15
Now he wouldn't try to prove that by himself... :) –  Salech Alhasov Dec 9 '13 at 13:17
Certainly he should; ideally you should at least try to think of a proof of any fact you're presented before "peeking" at the textbook/canonical solution. However, I thought it was important to give the name because a) this is a good "trick" to be able to reference when needed and b) the usual summation-by-parts proof is a surprising approach, and is worth seeing. –  Omnomnomnom Dec 9 '13 at 13:26
I'm glad that you did that! Cheers! –  Salech Alhasov Dec 9 '13 at 13:33
@Omnomnomnom I don't think you need to invoke Abel's test here. If the $a_n$ were only conditionally convergent, I would agree, but here they're absolutely convergent, so you can just crash through with the triangle inequality. –  Potato Dec 9 '13 at 18:21

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.

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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?

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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.
For $-1 < a_n < 0$, you have $\lvert a_n\rvert < \dfrac{\lvert a_n\rvert}{\lvert 1+a_n\rvert}$. –  Daniel Fischer Dec 9 '13 at 13:08