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I believe I found a proof for the divergence of this sum for any value of $z$ besides 0.

We can look on the telescopic series: $$\sum_{n=1}^{\infty}z^{1/(n+1)}-z^{1/n} = \lim_{N\rightarrow \infty} \left(z^{1/(N+1)}-z\right) = 1-z$$

If the sum in the title were convergent, as in equals, S, then we should have:

$$ S-z - S = 1-z \Rightarrow 1=0$$

So from this contradcition it follows that the series is divergent.

Is there any meaningful thing to say on this series? perhaps asymptotic analysis on this?

Thanks in advance.

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Perhaps you could add half a line to explain what "for any value of $\,z\,$ besides zero" means: if $\,0<z\in\Bbb R\,$ then everything's fine, as did pointed out in his answer; otherwise you're going to have to mess with complex numbers, branches of $\,n$-th root functions and stuff. –  DonAntonio Jun 4 '13 at 10:27
    
First I thought of the real case, how would the general complex will differ from the real case? –  MathematicalPhysicist Jun 4 '13 at 12:23
    
Well, in fact it doesn't since $\,z=|z|e^{i\theta}\;,\;\;\theta\in\Bbb R\,$ and things get nice no matter what branch of the $\,n$-th root you choose, yet I think this may be worthwhile mentioning... –  DonAntonio Jun 4 '13 at 12:33

1 Answer 1

up vote 6 down vote accepted

The reasoning in your post is based on (and partially rediscovering) the fact that if $x_n\to x\ne0$ then the series $\sum\limits_nx_n$ diverges.

A more direct approach to this result is to note that if $\sum\limits_nx_n$ converges then $\sum\limits_{n\leqslant N}x_n$ and $\sum\limits_{n\leqslant N-1}x_n$ both converge to a same limit $\ell$ hence their difference $x_N$ converges to $\ell-\ell=0$.

In your context, $x_n=z^{1/n}$ hence $x_n\to x=1\ne0$ for every $z\ne0$ hence the series $\sum\limits_nz^{1/n}$ diverges. Asymptotics of $x_n$ when $n\to\infty$ are $x_n=1+\frac1n\log z+O\left(\frac1{n^2}\right)$ hence $$ \sum\limits_{1\leqslant n\leqslant N}z^{1/n}=N+\log N\cdot\log z+O(1). $$ Even more precisely, the sequence of general term $\left(\sum\limits_{1\leqslant n\leqslant N}z^{1/n}\right)-N-\log N\cdot\log z$ converges to a finite limit.

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Do we have any further meaningful things to say on this series? I mean asymptotically. –  MathematicalPhysicist Jun 4 '13 at 10:51
    
See revised version. –  Did Jun 4 '13 at 11:32

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