TITLE: Proof that $\sum\limits_{n=1}^{\infty} z^{1/n}$ does not converge
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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.