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?
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$$\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} \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}
$\endgroup$ $\begingroup$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.
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