Show that the series $\displaystyle{\sum_{n=1}^{\infty} \left(\frac{x}{n} \ln(1 + \frac{x}{n})\right) }$ converges uniformly on $(-1, A)$ for $A > -1$ and that the sum of the series has derivatives of all orders on $(-1,\infty)$.
Attempt: (a) I cannot bound the series. I did $\sum_{n=1}^{\infty}(\ln(n+x)-\ln(n))^{\frac{x}{n}}$. Then what next?
Hint: $\lim_{x\to0}\dfrac{\ln(1+x)}{x}=1$. In particular, there is an interval $[-a,a]$, $0<a<1$, such that $|\ln(1+x)|\le2\,|x|$ if $x\in[-a,a]$.