Does the following series converges or diverges: $$ \sum_{n=1}^\infty\left(\frac{1}{n} - \ln\left(\frac{n+1}{n}\right)\right) $$
I know that $ \sum_{n=1}^\infty \frac{1}{n} $ diverges and my intuition is that $\ln\frac{n+1}{n}$ doesnt effect it as it is going to $0$ as $n\to\infty$ so it still diverges but I'm not sure how to show it.
Thanks
First note that $$\sum_{n=1}^k(\frac{1}{n} -\log(n+1)+\log(n))) =\sum_{n=1}^k\frac{1}{n}-\log(k+1) =\log k+\gamma+\epsilon_k-\log(k+1) $$ since the series telescopes (where $\epsilon_k\to 0$ and $\gamma$ is the Euler-Mascheroni constant). Finally $$ \log k+\gamma+\epsilon_k-\log(k+1)=\log\left(\frac{k}{k+1}\right)+\gamma+\epsilon_k\to\gamma $$ as $k\to \infty$ so that the series converges.
Use direct comparison test:
$$0<\frac1n-\ln\left(\frac{n+1}n\right)<\frac1{2n^2}$$
which follows from the Taylor expansion of the natural logarithm.
By Frullani's theorem $$ \log\frac{n+1}{n}=\int_{0}^{+\infty}\frac{e^{-nx}-e^{-(n+1)x}}{x}\,dx \tag{1}$$ and since $\frac{1}{n}=\int_{0}^{+\infty}e^{-nx}\,dx = \frac{1}{n}$, by the dominated convergence theorem we get $$\begin{eqnarray*} \sum_{n\geq 1}\left(\frac{1}{n}-\log\frac{n+1}{n}\right) &=& \int_{0}^{+\infty}\left(1-\frac{1-e^{-x}}{x}\right)\sum_{n\geq 1}e^{-nx}\,dx\\&=&\int_{0}^{+\infty}\left(1-\frac{1-e^{-x}}{x}\right)\frac{dx}{e^x-1}\\&=&\int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx \tag{2}\end{eqnarray*}$$ where the function $g(z)=\frac{1}{e^z-1}-\frac{1}{ze^z}$ is regular in a right neighbourhood of the origin and has an exponential decay as $z\to +\infty$, hence it belongs for sure to $L^1(\mathbb{R}^+)$, implying that the original series is convergent. Additionally, the inequality $$ \frac{1}{2}e^{-z}\leq \frac{1}{e^z-1}-\frac{1}{ze^z} \leq \left(\frac{1}{2}+\frac{z}{12}\right)e^{-z} \tag{3}$$ implies that $$ \sum_{n\geq 1}\left(\frac{1}{n}-\log\frac{n+1}{n}\right)\in \left[\frac{1}{2},\frac{7}{12}\right].\tag{4}$$
The sequence of partial sums: $S_n = 1+\dfrac{1}{2} + \dfrac{1}{3}+ \cdots+ \dfrac{1}{n} - \ln n - \ln\left(\frac{n+1}{n}\right) \to \gamma$ ( the Euler constant ) which is the value of the series.
