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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 ?

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Sure, take any convergent series with nonzero terms and prepend it's negated sum as the first term. – Marcin Łoś Apr 29 '14 at 22:59
@MarcinŁoś Great observation. I was looking to show that the behaviour could be what I wanted, rather than for simplicity ... – Mark Bennet Apr 29 '14 at 23:02
up vote 5 down vote accepted

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.

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the limit of this series is log(2). proof – Michael Apr 29 '14 at 23:10
@Michael: I think you overlooked that this isn't the alternating harmonic series (for example) but one in which odd and even consecutive terms cancel exactly, – hardmath Apr 29 '14 at 23:20
yes, you're right! – Michael Apr 29 '14 at 23:24

How about $-1+\displaystyle \sum_{n=1}^{\infty} \left(\dfrac{1}{2^n}\right)$?

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no... it diverges to $-\infty$ – Michael Apr 29 '14 at 23:07
Every term is negative - so this isn't what you were thinking of? – Mark Bennet Apr 29 '14 at 23:07
@Michael OOps...that's embrassing – TTY Apr 29 '14 at 23:08
hmm, that's better. it wasn't me who downvoted your answer, btw. – Michael Apr 29 '14 at 23:17
@Michael doesn't matter, that's my first down vote, I like it. – TTY Apr 29 '14 at 23:20

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...

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$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 ...

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Note - I chose some illustrations which illustrate some possibilities of different behaviour (you can make the asymptotics as slow as you like, and have rapidly increasing numbers of negative terms between positive terms). So what do you mean in the question by "what can be said"? – Mark Bennet Apr 29 '14 at 23:10

How about

$$\sum_{n=1}^\infty \left( \frac{-1^n}{n^2} + \frac{\pi^2}{12\cdot 2^n} \right)$$

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Remember series for $\cos x$? Guess what happens when you put $x=\frac{\pi}2$?

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