finding region for uniformly convergent complex series

Question

Question:

$\sum_{n=0}^{\infty}z^2/(1+z^2)^n$ this is the series given for the question. In what region is this series uniformly convergent?

My attempt:

$|z^2/(1+z^2)^n| = |(z^2/(1+z^2)) * 1/(1+z^2)^{n-1}| < |(z^2/(1+z^2))| * |1/(1+z^2)^{n-1}| < |1/(1+z^2)^{n-1}|$

in order for $|1/(1+z^2)^{n-1}|$ to be convergent, $|1+z^2|>1$ should be the region for uniform convergence (Consider Weierstrauss M- test)

However, the answer, if the solution is right, is said to be $|1+z^2|>= R >1$

Is there a part in my solution where the inequality is not correct, or something I missed? I don't really get why $R$ came out when I think it is unnecessary.

Thanks in advance.

Answer 1

The Weierstrass $M$-test says that if there is a convergent series $\sum_{n=0}^\infty M_n$ such that, for each $z$ in a region $R$, you have $|f_n(z)|\leqslant M_n$, then the series $\sum_{n=0}^\infty f_n$ converges uniformly. So, simply knowing that you always have $|f_n(z)|<1$ is not enough. But if $|1+z^2|\geqslant R>1$, then, by your computations (which I did not check), you always have$$\left|\frac{z^2}{(1+z^2)^n}\right|\leqslant\left(\frac 1R\right)^n.$$So, since the series $\sum_{n=0}^\infty(1/R)^n$ converges, your series converges uniformly.

Answer 2

You may apply the $n$-root criteria. $$\Big(\Big|\frac{z^2}{(1+z^2)^n}\Big|\Big)^{1/n}=\frac{|z|^{2/n}}{|1+z^2|}\xrightarrow{n\rightarrow\infty}L(z)$$ where $L(0)=0$ or $L(z)=\frac{1}{|1+z^2|}$ for $z\neq0$. Uniform convergence occurs for all $z$ such that $L(z)\leq c<1$. for example, of $z=0$ or $|1+z^2|>\frac{1}{c}$ for $0<c<1$.