Why does arctan(1/n) summed not converge? My question is as follows: WolframAlpha tells me that the sum series S doesn't converge, but why? 
$$S=\sum_{n=1}^{\infty}\arctan(\frac{1}{n})$$
$$\lim_{n\to \infty} \arctan(\frac{1}{n})=0$$
So S (slowly) stops growing when n gets larger and larger. So why doesn't it converge? Shouldn't it stop growing near infinity, making it converge?
 A: Since $\arctan(x)\geq\frac{\pi x}{4}$ for $x\in[0,1]$ ($\arctan(x)$ is concave for $x\geq 0$), it follows that
$$\sum_{n=1}^{\infty}\arctan\left(\frac{1}{n}\right)\geq\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{1}{n}=+\infty.$$
where in the last step we used the fact that the Harmonic series is divergent.
A: For values of $x$ near $0$, $\arctan(x)$ ~ $x$.
IF you know calculus, this is because the rate of change of $\arctan(x)$, $\displaystyle \frac{1}{1+x^2}$, approaches $1$,the rate of change of $x$, as $x$ approaches $0$.
It is a well known fact that the harmonic series or $\displaystyle \frac{1}{x}$, that is $1+\frac{1}{2}+\frac{1}{3}...$ does not converge.
For $\arctan{\frac{1}{x}}$, as $x$ gets bigger, this series slowly starts to become the harmonic series, which diverges.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\left.\arctan\pars{1 \over n}\right\vert_{\ n\ \geq\ 1} & = {1 \over n}\,\left.1 \over \xi^{2} + 1\right\vert_{\ 0\ <\ \xi\ <\ 1/n} > {1 \over n}\,{1 \over \pars{1/n}^{2} + 1} = {n \over n^{2} + 1} > {n \over n^{2} + n^{2}} = {1 \over 2}\,{1 \over n}
\end{align}
A: For the same reason that the harmonic series $\sum_n 1/n$ diverges. Indeed
$\arctan(1/n)\sim 1/n$ as $n\to\infty$, so your series diverges by
limit-comparison with the harmonic series.
