# Math Help - Uniform convergence

1. ## Uniform convergence

1) $Let~M\in\mathbb{R}>0$ Show that $\sum_{n=1}^{\infty} \frac{z^{2n}}{n!}$ converges uniformly for $|z| < M$.

That part I feel confident on. I use ratio test on $\sum_{n=1}^{\infty} \frac{M^{2n}}{n!}$ and find convergence, then use Weierstrauss M-Test to show the statement as required.

2) Does $\sum_{n=1}^{\infty} \frac{z^{2n}}{n!}$ converge uniformly on $\mathbb{C}$?

Here is where I'm a bit confused. Part 1 shows uniform convergence for |z| < M and doesn't set any requirements on M, but the only example I have says the answer is still no, as z=0 is convergence, but not uniformly. A bit of explanation would be wonderful! Thanks

2. ## Re: Uniform convergence

For part 2), remember the expansion for $e^w=\sum w^n/n!$.

3. ## Re: Uniform convergence

Originally Posted by Thomas154321
1) $Let~M\in\mathbb{R}>0$ Show that $\sum_{n=1}^{\infty} \frac{z^{2n}}{n!}$ converges uniformly for $|z| < M$.

That part I feel confident on. I use ratio test on $\sum_{n=1}^{\infty} \frac{M^{2n}}{n!}$ and find convergence, then use Weierstrauss M-Test to show the statement as required.

2) Does $\sum_{n=1}^{\infty} \frac{z^{2n}}{n!}$ converge uniformly on $\mathbb{C}$?

Here is where I'm a bit confused. Part 1 shows uniform convergence for |z| < M and doesn't set any requirements on M, but the only example I have says the answer is still no, as z=0 is convergence, but not uniformly. A bit of explanation would be wonderful! Thanks
Just use the ratio test for part 2.

The series is convergent wherever \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}, so solve

\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{\frac{z^{2 \left( n + 1 \right) } }{\left( n + 1 \right) !}}{\frac{z^{2n}}{n!}} \right| &< 1 \\ \lim_{n \to \infty} \left| \frac{n!\,z^{2n + 2}}{\left( n + 1 \right) n! \, z^{2n}} \right| &< 1 \\ \lim_{n \to \infty} \left| \frac{z^2}{n + 1} \right| &< 1 \\ \left| z^2 \right| \lim_{n \to \infty} \frac{1}{n + 1} &< 1 \\ \left| z \right| ^2 \cdot 0 &< 1 \\ 0 &< 1 \end{align*}

The limit of the ratios is 0, which is ALWAYS less than 1, no matter what value of z. So that means that the series is convergent for all z.