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: $$\gamma= 1-\int_{1}^{\infty}{\frac{\{t\}}{t^2}dt}\approx 0.57721$$

Answer 2

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.

Answer 3

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.