Nature of infinite series $ \sum\limits_{n\geq 1}\left[\frac{1}{n} - \log(1 + \frac{1}{n})\right] $

Question

$$\sum\limits_{n\geq 1}\left[\frac{1}{n} - \log\left(1 + \frac{1}{n}\right)\right]$$ Is it convergent or divergent? Wolfram suggests to use comparison test but I can't find an auxiliary series.

Answer 1

We have that

$$\frac{1}{n} - \log\left(1 + \frac{1}{n}\right)= \frac{1}{n}-\frac{1}{n}+\frac{1}{2n^2}+O\left(\frac1{n^3}\right)=\frac{1}{2n^2}+O\left(\frac1{n^3}\right)$$

therefore the given series converges by limit comparison test with $\sum \frac 1{n^2}$.

As an alternative, since $\log (1+x)\ge x-\frac12 x^2$ we have

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

and therefore the given series converges by comparison test with $\sum \frac 1{2n^2}$.

Answer 2

We may take the series $\sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]$ as an equivalent definition of the Euler-Mascheroni constant $\gamma=\lim_{n\to +\infty}\left(H_n-\log n\right)$, where $H_n=\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{n}$ is the $n$-th harmonic number.

Over the interval $(0,1)$ we have that $\frac{x-\log(1+x)}{x^2}$ is a decreasing function, going from $\frac{1}{2}$ to $1-\log(2)$. In particular $\gamma$ is bounded between $(1-\log 2)\frac{\pi^2}{6}$ and $\frac{\pi^2}{12}$. More accurate approximations can be derived from creative telescoping, the integral representation

$$ \gamma=\int_{0}^{1}-\log(-\log x)\,dx$$ or the Shafer-Fink inequality. Actually $\gamma$ is pretty close to $\frac{1}{\sqrt{3}}$.
The irrationality of $\gamma$ is a long-standing open problem.

Answer 3

By MVT applied to $$f:x\mapsto x-\ln(1+x)$$ in $[0,1/n]$,

with $$f'(x)=\frac{x}{1+x}\le x\le \frac 1n$$ $$f(\frac 1n)=\frac 1n f'(c)\le \frac{1}{n^2}$$

by comparison test, the series converges.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffb]{\ds{\sum_{n \geq 1}\bracks{{1 \over n} - \ln\pars{1 + {1 \over n}}}}} \\[5mm] & =\ \overbrace{\lim_{N \to \infty}\bracks{\sum_{n = 1}^{N}{1 \over n} - \ln\pars{N}}}^{\ds{\stackrel{\mrm{def.}}{=}\ \gamma}} \\[2mm] & + \lim_{N \to \infty}\bracks{\ln\pars{N} - \sum_{n = 1}^{N}\ln\pars{n + 1} + \sum_{n = 1}^{N}\ln\pars{n}} \\[5mm] & = \gamma + \lim_{N \to \infty}\bracks{\ln\pars{N} - \sum_{n = 2}^{N + 1}\ln\pars{n} + \sum_{n = 1}^{N}\ln\pars{n}} \\[5mm] & = \gamma + \lim_{N \to \infty}\braces{\!\!\ln\pars{N} - \bracks{\sum_{n = 2}^{N}\ln\pars{n} + \ln\pars{N + 1}} + \sum_{n = 2}^{N}\ln\pars{n}\!\!} \\[5mm] & = \gamma + \lim_{N \to \infty}\ln\pars{1 \over 1 + 1/N} = \bbx{\gamma} \end{align}

$\ds{\gamma}$ is the Euler-Mascheroni Constant.

Answer 5

Using this inequality $$x\geq\ln{(1+x)}\geq \frac{x}{1+x}, \forall x>-1$$ we have $$0\leq\frac{1}{n}-\log{\left(1+\frac{1}{n}\right)}\leq \frac{1}{n}-\frac{1}{n+1}$$ and as a result $$0\leq \sum\limits_{n\geq1}\left[\frac{1}{n}-\log{\left(1+\frac{1}{n}\right)}\right]\leq \lim\limits_{n\rightarrow\infty}\left(1-\frac{1}{n+1}\right)=1$$