Convergent or divergent series examples Suppose $\sum a_n$  is convergent. Is $\sum {{a_n} \over {1+|a_n|}}$ convergent or divergent?
 A: If $\sum a_n$ converges absolutely, the the answer is affimative. We claim that this is no longer the case for conditional convergence. Note that
$$ \frac{x}{1+|x|} = x - x|x| + O(x^3)$$
near the origin. Now consider the series
$$\sum_{n=1}^{\infty} a_n = \frac{2}{\sqrt{1}} - \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{1}} + \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}} + \cdots.$$
This series converges conditionally. Now then we have
$$  \frac{a_{3n-2}}{1+|a_{3n-2}|} +  \frac{a_{3n-1}}{1+|a_{3n-1}|} +  \frac{a_{3n}}{1+|a_{3n}|} = -\frac{2}{n} + O\left( \frac{1}{n^{3/2}}\right). $$
Therefore the sum $\sum \frac{a_n}{1+|a_n|}$ diverges.
Slightly modifying this argument also generates a conditionally convergent series $\sum a_n$ whose corresponding sum $\sum \frac{a_n}{1+|a_n|}$ also converges, thus the answer is inconclusive.
A: Surprisingly, it need not be convergent.  In fact, if $f$ is a function such that $\sum_n a_n$ converges iff $\sum_n f(a_n)$ converges, then $f$ must be linear in some neighbourhood of $0$.
See https://groups.google.com/forum/?hl=en&fromgroups=#!topic/sci.math.research/SzZDXRFyvhk
