The series of function $f(x)=\sum_{n\geq 1}\frac{1}{n}\ln(1+\frac{x}{n})$; the convergence and the differentiability. 
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*}
 A: We have $$f(x)=\sum_{n\geq
 1}\frac{1}{n}\ln(1+\frac{x}{n}),\qquad x>-1. \tag1$$

a) Show that the series is pointwise convergent.

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 $\displaystyle \sum_{n\geq
 1}\frac{1}{n^3}$.

b) Determine the termwise differentiated series, and show that it is
  uniformly convergent.

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 $[\alpha,+\infty),\, -1<\alpha<1$.

c) Explain that $f$ is differentiable.

You may check that all hypotheses of theorem 2 (a standard theorem allowing termwise differentiation of a series) are satisfied. 
