I have a problem with the power series:
$$\sum_{n=1}^\infty \frac{1}{nx}(1-e^{-\frac{x}{n}})~~~~~~x\in(0, \infty)$$
I have to prove that it's uniformely convergent and determine whether the sum is differentiable.
If the series is defined by $f(x)$, if I find $f(x)<a_n $ and $\sum a_n $ is convergent, then $\sum f(x) $ is uniformely convergent (Weierstrass). To determine whether the sum is differentiable, I have to determine whether the sum of derivatives is convergent. However, how can I do that? Can I split it to:
$$\sum_{n=1}^\infty \frac{1}{nx} - \sum_{n=1}^\infty \frac{1}{nx}e^{-\frac{x}{n}} $$
Does it help anything?
No, because $\sum_{n=0}^\infty {1\over n}$ is $\infty$. Its derivative is $$f'(x){= \sum_{n=1}^\infty \left[-\frac{1}{nx^2}(1-e^{-\frac{x}{n}})+\frac{e^{-\frac{x}{n}}}{n^2x}\right] \\=-{f(x)\over x}+\sum_{n=1}^\infty \frac{e^{-\frac{x}{n}}}{n^2x} } $$ Note that the second limit is convergent compared to $\sum_{n=1}^\infty \frac{1}{n^2}$