Convergence of the series $\sum^{\infty}_{n=2} \left(\ln\left(\frac{n}{n-1}\right) - \frac{1}{n}\right) $ Does the following series converge or diverge?
\begin{equation}
 \sum^{\infty}_{n=2} \left(\ln\left(\frac{n}{n-1}\right) - \frac{1}{n}\right)
\end{equation}
I have noticed that each of partial sum telescopes leaving me with:
\begin{equation}
S_n = \ln(n) - \sum^{n}_{k=2}\frac{1}{k}
\end{equation}
I know that harmonic series are divergent, I am not sure how to use that fact in this case though. How should one follow from here?
 A: $$\log\left( \frac{n}{n-1}\right) -\frac{1}{n}=-\log\left(\frac{n-1}{n}\right)-\frac{1}{n}= -\log\left(1-\frac{1}{n}\right)-\frac{1}{n}$$

$$ -\log\left( 1-\frac{1}{n}\right)=\frac{1}{n}+\frac{1}{2n^2}+o\left( \frac{1}{n^2}\right)$$

$$\log\left(\frac{n}{n-1}\right)-\frac{1}{n} =\frac{1}{2n^2}+o\left( \frac{1}{n^2}\right)$$ The series is therefore convergent.
$$\underline{\textbf{About the limit of this sum}}:$$

Let $\gamma$ be the limit. Using partial summation Lemma:
$$\sum_{n\le x}{\frac{1}{n}}=\frac{\lfloor{x}\rfloor}{x}+\int_{1}^{x}{\frac{\lfloor{t}\rfloor}{t^2}dt}= \frac{\lfloor{x}\rfloor}{x}+\int_{1}^{x}{\frac{t-\{t\}}{t^2}dt} $$
$$= \frac{\lfloor{x}\rfloor}{x}+\log{x}-\int_{1}^{x}{\frac{\{t\}}{t^2}dt} $$ so:
$$\sum_{n\le x}{\frac{1}{n}}-\log{x}= \frac{x+O(1)}{x}-\int_{1}^{x}{\frac{\{t\}}{t^2}dt}$$
$$=1-\int_{1}^{\infty}{\frac{\{t\}}{t^2}dt}+ \underbrace{\int_{x}^{\infty}{\frac{\{t\}}{t^2}dt}}_{=O(\frac{1}{x})}+O(\frac{1}{x})$$
$$ \sum_{n\le x}{\frac{1}{n}}-\log{x}=\gamma+O(\frac{1}{x})$$ where: $$\gamma= 1-\int_{1}^{\infty}{\frac{\{t\}}{t^2}dt}\approx 0.57721$$

A: Notice that $$\ln\left(\frac{n}{n-1} \right) = - \ln\left(1-\frac1n\right) = -\left(-\frac 1 n - \frac 1 {2n^2} - \text{smaller terms}\right).$$ Thus $$\left( \ln\left(\frac{n}{n-1} \right) - \frac 1 n\right) \sim \frac 1 {2n^2}.$$ Thus the series converges by comparison with $\sum \frac 1 {n^2}$, and as pointed out in the comments, the series sums to $\gamma$, the Euler-Mascheroni constant.  
A: Using this inequality
$$x\geq\ln{(1+x)}\geq \frac{x}{1+x}, \forall x>-1 \tag{1}$$
and
$$\log{\left(\frac{n}{n-1}\right)}-\frac{1}{n}=
\log{\left(1+\frac{1}{n-1}\right)}-\frac{1}{n} \tag{2}$$
we have
$$0=\frac{1}{n}-\frac{1}{n}=
\frac{\frac{1}{n-1}}{1+\frac{1}{n-1}}-\frac{1}{n}\overset{(1)}{\leq}
\log{\left(1+\frac{1}{n-1}\right)}-\frac{1}{n}\overset{(1)}{\leq}
\frac{1}{n-1}-\frac{1}{n}$$
and as a result, from $(2)$
$$0\leq \sum\limits_{n\geq2}\left(\log{\left(\frac{n}{n-1}\right)}-\frac{1}{n}\right)\leq 
\sum\limits_{n\geq2}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\\
\lim\limits_{k\rightarrow\infty}\sum\limits_{n=2}^{k}\left(\frac{1}{n-1}-\frac{1}{n}\right)=
\lim\limits_{k\rightarrow\infty}\left(1-\frac{1}{k}\right)=1$$
and the original series converges.
