What can be said about a series with nonzero terms, whose sum is zero? What can be said about an infinite series $\sum^{\infty}_{n=1} a_n$ with $a_n \neq 0$ for all $n$, whose sum is zero ? Does such a series exist ? If yes, can you give an example ?
 A: Take any positive sequence $\{a_n\}$ that converges to zero, e.g. $a_n = 1/n$.  Define a series $\sum s_i$ whose even terms are $s_{2i} = -a_i$ and whose odd terms are $s_{2i-1} = a_i$.
The series is alternating and converges (always, but most easily by the alternating series test if monotonically decreasing).  It's easy to see the limit is zero since the even partial sums are identically zero.
A: How about $-1+\displaystyle \sum_{n=1}^{\infty} \left(\dfrac{1}{2^n}\right)$?
A: You can't really say that much about such a series at all. Clearly there have to be positive and negative terms if the series sums to zero and has non-zero terms. However it's possible to have such a series where any rearrangement of the terms gives any sum that you want, even divergent, so literally the series can just be anything as long as the positive terms eventually cancel out the negative terms, no matter what rearrangements of the series may give...
A: $1-1+\frac 12-\frac 12+\frac 14 -\frac 14+\frac 18-\frac 18\dots $ might be an example as would $1-\frac 12-\frac 12+\frac 12-\frac 14-\frac 14+\frac 13-\frac 16-\frac 16+\frac 14-\frac 18-\frac 18 \dots$
I am sure you can make up other examples ...
A: How about
$$\sum_{n=1}^\infty \left( \frac{-1^n}{n^2} + \frac{\pi^2}{12\cdot 2^n} \right)$$
A: Remember series for $\cos x$? Guess what happens when you put $x=\frac{\pi}2$?
