Convergence of the series $\sum^{\infty}_{n=2} \left(\ln\left(\frac{n}{n-1}\right) – \frac{1}{n}\right) $

Question

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?

Answer 1

$$\log\left( \frac{n}{n-1}\right) -\frac{1}{n}=-\log\left(\frac{n-1}{n}\right)-\frac{1}{n}= -\log\left(1-\frac{1}{n}\right)-\frac{1}{n}$$

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

$$\log\left(\frac{n}{n-1}\right)-\frac{1}{n} =\frac{1}{2n^2}+o\left( \frac{1}{n^2}\right)$$ 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: $$\gamma= 1-\int_{1}^{\infty}{\frac{\{t\}}{t^2}dt}\approx 0.57721$$