Is $\sum_{n=0}^{\infty}{\frac{2^{n}x^{n}}{n!}}$ convergent on $(0,1)$

Question

if anyone could help me with the following problem :

Let $\sum_{n=0}^{\infty}{\frac{2^{n}x^{n}}{n!}}$ ,

I have to see the pointwise convergence on $(0,1)$, I know that the uniform convergence implies it. Here is what I thought may help me but got me nowhere:

A series $\sum_{n=1}^{\infty}{f_{k}(x)}$ converges uniformly if the sequence of partial sums $S_{n}(x)$ = $\sum_{k=1}^{n}{f_{k}(x)}$ converges uniformly.

I thought also maybe using the fact that : $\left|f_{n}(x) \right| \leq M_{n} \; \text{s.t} \sum_{n=1}^{\infty}{M_{n}} < \infty $ (The Weierstrass M–test)

Or using the fact that : $\lim_{n\rightarrow \infty}(\sup\left|f_{n}(x) \right|) = 0$

I also thought maybe rewriting $\sum_{n=0}^{\infty}{\frac{2^{n}x^{n}}{n!}}$ as $\exp(2x)$ .. I'm kinda lost so if anyone could help it would be a lot appreciated.

Thanks in advance

Answer 1

The easiest way is to just use the ratio test. Pick some $x \in (0, 1)$. Then the ratio between consecutive terms is $$ \frac{\frac{2^nx^n}{n!}}{\frac{2^{n+1}x^{n+1}}{(n+1)!}} = \frac{2x}{n+1}.$$ It's easy to see that for any such $x$ we have $$ \lim_{n \to \infty} \frac{2x}{n+1} = 0 < 1$$ so this converges pointwise by the ratio test.