MHB How to prove the convergence of a series with absolute values?

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The discussion presents a problem of proving the convergence of a series involving absolute values, specifically focusing on the series $\sum a_n$ that is absolutely convergent. It states that if $\sum a_n$ is absolutely convergent, then the series $\sum \dfrac{a_n^2}{1 + a_n^2}$ also converges. Additionally, if $a_n \neq -1$ for all $n$, it asserts that $\sum \dfrac{a_n}{1 + a_n}$ converges absolutely. Despite the posed problem, no participants provided answers, and the original poster shares their solution afterward. The thread emphasizes the importance of understanding convergence in series involving absolute values.
Euge
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Happy New Year everyone! Here is this week's POTW:

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If $\sum a_n$ is an absolutely convergent series of nonzero real numbers, prove that $\sum \dfrac{a_n^2}{1 + a_n^2}$ converges. If, in addition, $a_n \neq -1$ for all $n$, show that $\sum \dfrac{a_n}{1 + a_n}$ converges absolutely.

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No one answered this week's problem. You can read my solution below.

By the estimate $1 + a_n^2 \ge 2|a_n|$, the series $\sum \dfrac{a_n^2}{1 + a_n^2}$ is dominated by the convergent series $\dfrac{1}{2}\sum |a_n|$. By the comparison test $\sum \dfrac{a_n^2}{1 + a_n^2}$ converges. If $a_n \neq -1$ for all $n$ as well, then using the fact that $|a_n| < 1/2$ for all sufficiently large $n$, it follows that $$\left|\frac{a_n}{1 + a_n}\right| \le \frac{|a_n|}{1 - |a_n|} < 2|a_n|$$ for $n$ sufficiently large. Since $\sum 2 |a_n| < \infty$, by comparison the series $\sum \dfrac{a_n}{1 + a_n}$ converges absolutely.
 

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