Uniform convergence of $\sum_{n=1}^\infty \frac{1}{z^n+1}$

Question

I've proved that the series $\sum_{n=1}^\infty \frac{1}{z^n+1}$ for $|z|\geq r >1,z\in\Bbb C$ converges uniformly by weierstrass M-test. But I want to show it's not uniformly convergent on $|z|>1$. Could you give any hints?

Answer 1

If it was uniformly convergent for $|z|>1$, $$ \mathop {\sup }\limits_{|z| > 1} \left|\sum\limits_{n = N}^\infty {\frac{1}{{z^n + 1}}} \right| $$ would tend to zero as $N\to +\infty$. Note however that $$ \mathop {\sup }\limits_{|z| > 1} \left|\sum\limits_{n = N}^\infty {\frac{1}{{z^n + 1}}} \right| \ge \sum\limits_{n = N}^\infty {\frac{1}{{\left( {1 + \frac{1}{N}} \right)^n + 1}}} \ge \frac{1}{{\left( {1 + \frac{1}{N}} \right)^N + 1}} \ge \frac{1}{{e + 1}}. $$

Answer 2

If it did, then it would converge uniformly on $(1,\infty)$. But each partial sum $\sum_{n=1}^N\frac1{z^n+1}$ is bounded there, and so, since we are assuming that the convergence is uniform, $\sum_{n=1}^\infty\frac1{z^n+1}$ would be bounded there too, since there is some $N\in\Bbb N$ such that$$(\forall n\in\Bbb N)(\forall z\in(0,\infty)):m\geqslant N\implies\left|\left(\sum_{n=1}^\infty\frac1{z^n+1}\right)-\left(\sum_{n=1}^m\frac1{z^n+1}\right)\right|<1,$$and this in turn, implies that$$(\forall z\in(0,\infty)):\left|\sum_{n=1}^\infty\frac1{z^n+1}\right|<1+\sup_{z\in(1,\infty)}\left|\sum_{n=1}^N\frac1{z^n+1}\right|.$$On the other hand, if $m\in\Bbb N$,$$\lim_{z\to1^+}\sum_{n=1}^m\frac1{z^n+1}=\sum_{n=1}^m\frac12=\frac m2,$$and therefore there is some $w_m\in(1,\infty)$ such that$$\sum_{n=1}^m\frac1{w_m^{\,n}+1}>\frac m4,$$from which it follows that$$\sum_{n=1}^\infty\frac1{w_m^{\,n}+1}>\frac m4.$$But then $\sum_{n=1}^\infty\frac1{z^n+1}$ is unbounded, and so a contradiction was reached.