Suppose $a_n$ are real numbers and $\sum a_n$ and $\sum a_n^2$ converges. How would one go about showing that $\sum \frac{a_n}{1+a_n}$ converges? ($a_n \neq -1$ for every $n$)
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Are we given that $a_n \neq -1$? – Ben Grossmann Apr 10 '15 at 20:21
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@Omnomnomnom, if $a_{n}=-1$ then neither $\sum a_{n}$ nor $\sum a_{n}^2$ converge. – TravisJ Apr 10 '15 at 20:23
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1@TravisJ I mean for any $n$ – Ben Grossmann Apr 10 '15 at 20:23
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If $a_n$ are positive, the problem isn't too hard... – Leonhardt von M Apr 10 '15 at 20:24
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@Omnomnomnom, sorry, that was my misunderstanding, I thought you meant a constant sequence (in retrospect it was silly to think you may have meant that). – TravisJ Apr 10 '15 at 20:28
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If $a_n=-1$, then $\sum a_n^2$ won't converge. – Eugene Zhang Apr 10 '15 at 20:32
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I will assume that $a_n \neq -1$ for every $n$.
Since $\lim a_n = 0$, by ignoring finitely many terms, we may assume that $|a_n| \leq 1/2$ for every $n$. Now notice that
$$ \frac{a_n}{1+a_n} = a_n - b_n,\qquad \text{with}\ b_n=\frac{a_n^2}{1+a_n}. $$
Since $|b_n| \leq 2 a_n^2$, the series $\sum b_n$ converges absolutely by comparison. This proves the convergence of the series $\sum a_n/(1+a_n)$ as well.
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Sangchul Lee
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(1) Since $\sum a_n$ is converge, then there exists $N$ s.t. $$ n \geq N \Rightarrow |a_n| \leq c < 1 $$ Fix $N$. We will assume that all $|a_n| \leq c$.
Hence $$ \sum \frac{a_n}{1+a_n} = \sum_n \sum_{i=1} (-1)^{i-1} a_n^{i} $$
(2) $$ |a_n^i| \leq |a_n|^2 |a_n|^{i-2} \leq |a_n|^2,\ i\geq 3 $$
Hence $\sum_n a_n^i$ is absolutely convergent for $i\geq 3$.
(3) Set $$ z_{in} = (-1)^{i-1} a_n^i $$
Then $$ x_n =\lim_i z_{in} =0,\ y_i=\lim_n z_{in} =0 $$
In further there exists $M$ : $$ i>M\Rightarrow |z_{in}- x_n| < c^i < c^M < \varepsilon $$
(4) By Iterated Limit Theorem, the iterated limit exists.
HK Lee
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