Nature of infinite series $ \sum\limits_{n\geq 1}\left[\frac{1}{n} - \log(1 + \frac{1}{n})\right] $ $$\sum\limits_{n\geq 1}\left[\frac{1}{n} - \log\left(1 + \frac{1}{n}\right)\right]$$
Is it convergent or divergent?
Wolfram suggests to use comparison test but I can't find an auxiliary series.
 A: 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 $\sum \frac 1{2n^2}$.
A: 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.
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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.

A: 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
$$0\leq \sum\limits_{n\geq1}\left[\frac{1}{n}-\log{\left(1+\frac{1}{n}\right)}\right]\leq \lim\limits_{n\rightarrow\infty}\left(1-\frac{1}{n+1}\right)=1$$
