Radius of convergence, and convergence of derivatives.

Question

I was looking at a simple exercise, but I have a doubt. I have to prove that this series converges uniformly in the region of convergence, and . $$ \sum_{n=0}^{\infty}\frac{z^n}{a^2+n^2} \quad a>0 $$ So it was easy to see that the radius of convergence is actually one. If i take the Dirichlet Abel test I would get that in all $z$ such that: $arg(z)\neq 2k\pi$ it converges. And by testing separately the case $z=1$ it also converges. So if it converges absolutely when $|z_1|=1$ then I got that for every $z$ in the closed region, it converges uniformly. $$$$I don't get the implication of the conditions on a. They ask me if the series that you get deriving term by term is uniform convergent on the closed region. Since this is: $$ \sum_{n=1}^{\infty}\frac{nz^{n-1}}{a^2+n^2} < \sum_{n=1}^{\infty}\frac{z^{n-1}}{n} $$ I know that this last series converges for all $z$ inside the unit circle and on its contour excepting $1$. So does this assures me that it doesn't converges uniformly in any $z$ such that $|z|=1$? $$$$Thanks a lot.

Answer 1

You get uniform convergence of $\sum_{n \geqslant 1} \frac{n z^{n-1}}{a^2 + n^2}$ on any compact subset of $D= \{z \in \mathbb{C} : |z|< 1\}$ using the Weierstrass test.

For example, if $|z| \leqslant r < 1$ then

$$\left|\frac{n z^{n-1}}{a^2 + n^2} \right| \leqslant \frac{r^{n-1}}{n},$$

and uniform convergence follows, since $\sum_{n \geqslant1 } \frac{r^{n-1}}{n}$ converges when $0 \leqslant r < 1$.

Convergence is not uniform even on the open disk $D$.

Note that for real $z \in [0,1)$ and all $n > a$ we have

$$\sum_{k=n+1}^{2n}\frac{k z^{k-1}}{a^2 + k^2} \geqslant n\frac{nz^{2n-1}}{a^2 + 4n^2} \geqslant \frac{z^{2n}}{4+ a^2/n^2 } \geqslant \frac{z^{2n}}{5 }.$$

Choosing a sequence $z_n = 1-1/n \in [0,1)$, we have

$$\lim_{n \to \infty}\sum_{k=n+1}^{2n}\frac{k z_n^{k-1}}{a^2 + k^2} \geqslant \lim_{n \to \infty}\frac{(1-1/n)^{2n}}{5} = \frac{e^2}{5} \neq 0,$$

which implies partial sums cannot satisfy the Cauchy criterion uniformly.