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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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    $\begingroup$ Sure, take any convergent series with nonzero terms and prepend it's negated sum as the first term. $\endgroup$ Apr 29, 2014 at 22:59
  • $\begingroup$ @MarcinŁoś Great observation. I was looking to show that the behaviour could be what I wanted, rather than for simplicity ... $\endgroup$ Apr 29, 2014 at 23:02

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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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  • $\begingroup$ the limit of this series is log(2). proof $\endgroup$
    – Michael
    Apr 29, 2014 at 23:10
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    $\begingroup$ @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, $\endgroup$
    – hardmath
    Apr 29, 2014 at 23:20
  • $\begingroup$ yes, you're right! $\endgroup$
    – Michael
    Apr 29, 2014 at 23:24
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How about $-1+\displaystyle \sum_{n=1}^{\infty} \left(\dfrac{1}{2^n}\right)$?

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  • $\begingroup$ no... it diverges to $-\infty$ $\endgroup$
    – Michael
    Apr 29, 2014 at 23:07
  • $\begingroup$ Every term is negative - so this isn't what you were thinking of? $\endgroup$ Apr 29, 2014 at 23:07
  • $\begingroup$ @Michael OOps...that's embrassing $\endgroup$
    – TTY
    Apr 29, 2014 at 23:08
  • $\begingroup$ hmm, that's better. it wasn't me who downvoted your answer, btw. $\endgroup$
    – Michael
    Apr 29, 2014 at 23:17
  • $\begingroup$ @Michael doesn't matter, that's my first down vote, I like it. $\endgroup$
    – TTY
    Apr 29, 2014 at 23:20
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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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  • $\begingroup$ 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"? $\endgroup$ Apr 29, 2014 at 23:10
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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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