Consider the series of function $f(x)=\sum_{n\geq 1}\frac{1}{n}\ln(1+\frac{x}{n})$ for $x>-1$.
a) Show that the series is pointwise convergent.
Answer: I actually don't know how to show it. I could show that the series is convergent for all $x>-1$. First I noticed that \begin{equation*} \frac{1}{n}\ln\left ( 1+\frac{x}{n} \right )=\ln\left ( \left ( 1+\frac{x}{n} \right )^{1/n} \right )<\left ( 1+\frac{x}{n} \right )^{1/n} \end{equation*} for all $x>-1$. Comparing the series with $\sum_{n\geq 1}\left ( 1+\frac{x}{n} \right )^{1/n}$ that diverges makes me to find other comparisons. I have not yet found a good comparison that converges. Or, if $\epsilon>0$ is given, then I am unable to find a $K>0$ such that \begin{equation*} \left | \sum_{n= 1}^{k}\frac{1}{n}\ln\left ( 1+\frac{x}{n} \right )-f(x) \right |<\epsilon \end{equation*} for all $k>K$.
b) Determine the termwise differentiated series, and show that it is uniformly convergent.
Answer: Let $f_{n}(x)=\frac{1}{n}\ln\left (1+\frac{x}{n} \right )$. Since $f_{n}(x)$ is differentiable with $f_{n}'(x)=\frac{1}{n(n+x)}$ for $x>-1$, so the termwise differentiated series is \begin{equation*} \left ( \sum_{n\geq 1} f_{n}(x) \right )'= \sum_{n\geq 1}f_{n}'(x)=\sum_{n\geq 1}\frac{1}{n(n+x)}. \end{equation*} Since, for example, $0\in (-1,\infty)$ and $\sum_{n\geq 1}f_{n}'(0)$ is convergent, then the series $\sum_{n\geq 1}f_{n}'(x)$ is uniformly convergent for all $x\in (-1,\infty)$.
c) Explain that $f$ is differentiable.
Answer: Isn't the problem b) enough to say that it is differentiable? Or, if $\epsilon>0$ is given, then I am unable to find a $\delta>0$ such that for all $x\in (-1,\infty)$ \begin{equation*} \left | \frac{f(x)-f(a)}{x-a}-f'(a) \right |<\epsilon\implies \left | x-a \right |<\delta. \end{equation*}
We have $$f(x)=\sum_{n\geq 1}\frac{1}{n}\ln(1+\frac{x}{n}),\qquad x>-1. \tag1$$
We may write, for any $x>-1$, as $n \to +\infty$, $$ \frac{1}{n}\ln\left ( 1+\frac{x}{n} \right )=\frac{1}{n}\left ( \frac{x}{n}+\mathcal{O}\left(\frac{x^2}{n^2}\right) \right )=\frac{x}{n^2}+\mathcal{O}\left(\frac{x^2}{n^3}\right) $$ and our series $(1)$ is pointwise convergent by comparison with $\displaystyle \sum_{n\geq 1}\frac{1}{n^2}$ and $\displaystyle \sum_{n\geq 1}\frac{1}{n^3}$.
We have $$ \sum_{n\geq 1}f_{n}'(x)=\sum_{n\geq 1}\dfrac{1}{n(n+x)},\qquad x>-1. \tag2 $$
For any $x \in [\alpha,+\infty),\, -1<\alpha<1$, we have $$ \left|\sum_{n\geq 1}\dfrac{1}{n(n+x)}\right| \leq \sum_{n\geq 1}\dfrac{1}{n(n-|\alpha|)} \tag3 $$ and the series $(2)$ is normally convergent thus uniformly convergent on $[\alpha,+\infty),\, -1<\alpha<1$.
You may check that all hypotheses of theorem 2 (a standard theorem allowing termwise differentiation of a series) are satisfied.