We have a first premilinary question that is: let $u_{n}$ be a sequence that converges to $u$, show that $$\lim_{n \to +\infty} \frac {e^{\frac {u_{n}} {n}} - 1} {\ln(1 + 1/n)} = u$$ this is easy with simple equivalents, but I am stumped at the second question.
Only thing I have so far is:
$$\lim_{n \to +\infty }\frac{e^{u_{1}}+e^{\frac {u_{2}}{2}}+e^{\frac {u_{3}}{3}} + ... + e^{\frac {u_{n}}{n}}-n}{\ln(n+1)} = \lim_{n \to +\infty }\frac{(e^{u_{1}} - 1) +(e^{\frac {u_{2}}{2}} -1) + e^{\frac {u_{n}}{n}}-1}{\ln(n+1)}$$
Which makes me think that I could use the Bolzano-Césaro theorem, but I can't manage to convert $\ln(n+1)$ to something useful. Thank you for your help.
Notice that $\ln(n+1)$ is increasing and unbounded. Furthermore:
$$ \frac{\sum_{i=1}^{n+1}e^{\frac{u_{i}}{i}}-(n+1)-\sum_{i=1}^{n}e^{ \frac{u_{i}}{i}}+n}{\ln(n+2)-\ln(n+1)} = \frac{\exp(\frac{u_{n+1}}{n+1})-1}{\ln(\frac{n+2}{n+1})} \rightarrow u $$
And thus the desired limit is $u$ in view of Cesaro-Stolz and the first part.