Proof of Convergence of a Series

• Jun 30th 2010, 02:45 PM
ragnar
Proof of Convergence of a Series
I'm trying to prove that, if $\displaystyle |z| > 1$ for any complex $\displaystyle z$, then $\displaystyle \displaystyle \sum^{\infty}_{n = 1} \frac{1}{1 + z^{n}}$ converges. I got this far but no farther:

If we try by comparison test, $\displaystyle |\frac{1}{1 + z^{n}}| \leq \frac{1}{|z|^{n} - 1} = \frac{1}{|z|^{n}} \cdot \frac{1}{1 - \frac{1}{|z|^{n}}}$. I know the first part of the RHS is decreasing, limit goes to 0. I'll be done if I show that the second part has bounded partial sums, but I can't figure that part out.

Thanks.
• Jun 30th 2010, 08:42 PM
theodds
How about the ratio test? :)
• Jul 1st 2010, 05:14 AM
ragnar
Nope, not unless you see something I don't. In any case, the final exam is upon me, so I won't need to know this in about three hours anyway.
• Jul 1st 2010, 06:22 AM
theodds
If $\displaystyle |z| > 1$ you should be able to show

$\displaystyle \displaystyle \frac{1 + z^n}{1 + z^{n + 1}} \to \frac{1}{z}$

since

$\displaystyle \displaystyle \left| \frac{1 + z^n}{1 + z^{n + 1}} - \frac{1}{z}\right| = \left|\frac{z - 1}{z(1 + z^{n + 1})} \right| \le \frac{|z| + 1}{|z| (|z|^n - 1)}$

and you can make $\displaystyle |z|^n$ as big as you like.