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


*

*Is my proof correct

 A: One other way to see this is to note that
$$\frac{\ln(n+1)-\ln n}{\ln n}\ge \int_{\ln n}^{\ln(n+1)}\frac{dt}{t}$$
hence
$$\sum_{n=2}^m\frac{\ln(n+1)-\ln n}{\ln n}\ge\int_{\ln 2}^{\ln(m+1)}\frac{dt}{t}
>\ln(\ln(m+1))$$
and the conclusion follows.
A: 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 $n$.
A: A faster way of getting to the same series to apply the Limit Comparison Test uses the mean value theorem:
$$\ln(1+n) - \ln(n) \approx \frac{1}{n}$$
$$\frac{\ln(1+n)}{\ln(n)} - 1 \approx \frac{1}{n\ln n}.$$
A: There have already been several solid ways forward presented.  Here we use standard inequalities only. 
To that end, we have the inequalities
$$\frac1{2n}<\frac{1}{n+1}\le\log\left(1+\frac1n\right) \tag 1$$
valid for $n\ge 2$.
Therefore, for $n\ge 2$, we have that
$$\frac{\log (n+1)-\log n}{\log n}\ge \frac{1}{(n+1)\log n} \ge \frac{1}{2n\log n} \tag 3$$
Inasmuch as the series on the right-hand side of $(3)$ diverges by the integral test, then the original series also diverges.  And we are done.
A: Note that
\begin{align}
\frac{\log(1+x)-\log(x)}{\log(x)}
\end{align}
is monotonically decreasing and positive when $x>1$. Applying Cauchy's condensation test, it suffices to look at
\begin{align}
\sum^\infty_{k=1}2^k\left( \frac{\log\left(1+\frac{1}{2^k}\right)}{k\log 2}\right).
\end{align}
Next, note that
\begin{align}
\frac{1}{2}x\leq \log(1+x)
\end{align}
for $0\leq x\leq 1$, then we see that
\begin{align}
\sum^\infty_{k=1}2^k\left( \frac{\log\left(1+\frac{1}{2^k}\right)}{k\log 2}\right) \geq \frac{1}{2\log 2}\sum^\infty_{k=1} \frac{1}{k}=\infty. 
\end{align}
Hence it follows that the original series also diverges. 
