Does $ \sum_{n\geq 2} \frac{\ln(1+n)}{\ln(n)}-1$ converge/diverge? How would you prove convergence/divergence of the following series?
$$ \sum_{n\geq 2}\left( \dfrac{\ln(1+n)}{\ln(n)}-1\right)$$
I'm interested in more ways of proving convergence/divergence for this series.
My thoughts
$$\dfrac{\ln(1+n)}{\ln(n)}=\frac{\ln(n(1+\dfrac{1}{n})}{\ln(n)}
=\frac{\ln(n)+\ln(1+\frac{1}{n})}{\ln(n)}
=1+\frac{\ln(1+\frac{1}{n})}{\ln(n)}$$
then 
$$\dfrac{\ln(1+n)}{\ln(n)}-1=\frac{\ln(1+\frac{1}{n})}{\ln(n)}$$
note that  $\ln(1+\frac 1n)=\frac 1n+o(\frac 1n)$ then $\ln(1+\frac 1n)\sim \frac 1n$ thus $u_n-1\sim \frac 1{n\ln(n)}$
or the serie $\dfrac{1}{n\ln(n)}$ divergent by Bertrand's test
the sum up $ \sum_{n\geq 2} \dfrac{\ln(1+n)}{\ln(n)}-1$ divergent


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*Is my proof correct


Also note, using the simple fact that $\lim_{n\to\infty} \left(1+1/n\right)^n =e$, that when $n$ large, we have 
$$\dfrac{\ln(1+1/n)}{\ln(n)}=\dfrac{\ln(1+1/n)^n}{n\ln(n)}\approx \frac{1}{n \ln(n)}$$
where for getting the equality I mutiplied both numerator and denominator by $n$.
