Prove series $\sum \frac{a_i}{1+|a_i|}$ converges If $\sum_{i=1}^n a_i$ converges, does $\sum_{i=1}^n \frac{a_i}{1+|a_i|}$ always converge? If not, please give a counter-example. Thanks.
 A: Here is a counterexample:
Let $c_n$ be a sequence of positive reals that converges slowly to $0$.  Consider the finite segment $S_n$ consisting of the $n+1$ $S_n := \{-c_n, \frac{c_n}{n}, \frac{c_n}{n}, \ldots, \frac{c_n}{n}\}$ summing to $0$.  If we concatenate these finite segments $S_n$ for $n=1$ to $\infty$, we get a series that converges to $0$, provided we ensure that $c_n \to 0$ so that partial sums actually tend to $0$ within each segment.  Other than this constraint we are free to choose $c_n$ arbitrarily.
Now let's look at the transformed form of $S_n$: the first term $-c_n$ maps to $-c_n/(1+c_n)$, and the other $n$ terms sum to $nc_n/(n+c_n)$.  These sum to $$ \frac{-c_n(n+c_n) + nc_n (1+c_n)}{(1+c_n)(n+c_n)} = \frac{(n-1)c_n^2}{(1+c_n)(n+c_n)} \sim c_n^2.$$
We can easily choose $c_n \to 0$ so that the sum of these diverges.  For instance $c_n = 1/\sqrt{n}$, or $c_n = 1/\log n$.
It looks like there's a lot of leeway to fiddle with this construction.  It doesn't appear critical to let the segment length grow to infinity, so maybe it can be simplified to a true alternating series.
