If $\sum a_n$ converges then $\sum (-1)^n \frac {a_n}{1+a_n^2}$ converges? Could you please give me some hint how to deal with this question:
If $\sum a_n$ converges,
does this necessarily mean that $\sum (-1)^n \frac {a_n}{1+a_n^2}$ must converge also ?
Thanks.
 A: Consider $a_n = (-1)^n\cdot \dfrac{1}{n}$, then the conclusion does not follow.
A: In fact there's a theorem: if $f$ is a function on $\mathbb R$ such that $\sum_n f(a_n)$ converges whenever $\sum_n a_n$ converges, then $f$ is linear in a neighbourhood of $0$.  This was proved by G. Wildenberg, American Mathematical Monthly 95 (1988)
542-544.  Y. Benjamini's solution to Problem E3404, American Mathematical
Monthly 99 (1992) 466-467 contains an extension:
Let $f: X \to Y$ be a mapping of normed spaces such that $\sum_{n=1}^\infty f(a_n)$ converges whenever $\sum_{n=1}^\infty a_n$ converges (both in the norm topology).
Then there is a neighbourhood of $0$ on which $f$ is equal to a bounded linear
operator.
I have a slight additional extension (unpublished):
Let $f: X -> Y$ be a mapping of Banach spaces such that $\sum_{n=1}^\infty f(a_n)$ converges weakly whenever $\sum_{n=1}^\infty a_n$ converges (strongly).  Then
there is a neighbourhood of $0$ on which $f$ is equal to a bounded linear
operator.
