$$f(x)= \sum \limits _{n=1}^\infty \frac{(-1)^{n}}{\sqrt{n}} e^{\frac{x}{n}} $$
(i) Show that for each $r > 0$, the series converges uniformly on $[−r,r]$
(ii) Is $f$ is differentiable on $\mathbb{R}$? Justify your answer.
This is what I have done so far.
ANSWER:
(i) Let $\ f_n(x)= \frac{(-1)^{n}}{\sqrt{n}} e^{\frac{x}{n}}. $ Then $f= \sum \limits _{n=1}^\infty f_n .$ Since $\frac{(-1)^{n}}{\sqrt{n}}$ converges by the alternating series test $\sum \limits _{n=1}^\infty f_n(0)$ converges.
Applying Weierstrass M-test to $f'_n$, we have $||f'_n = \frac{(-1)^{n}}{{n}\sqrt{n}} e^{\frac{x}{n}}||\le \frac{1}{{n}\sqrt{n}} $ Therefore $f'_n$ converges uniformly on $[-r,r]$.
Therefore $f= \sum \limits _{n=1}^\infty f_n $ converges uniformly on $[-r,r]$
(ii) By the above $f(x)$ is differentiable.