How to show this trig series is absolutely convergent: $\sum\limits_{n=7}^\infty \frac{\arctan(1/n)}{n}$? We have the following series: $$ \displaystyle \sum_{n=7}^\infty \dfrac{\arctan(1/n)}{n} $$
I want to show this series is absolutely convergent. Since $\arctan(1/x) > 0$ for every $x \in \mathbb{N}$, we know that $ \displaystyle \sum_{n=7}^\infty \dfrac{\arctan(1/n)}{n} =  \displaystyle \sum_{n=7}^\infty | \dfrac{\arctan(1/n)}{n} | $ and thus we only have to show it's convergent to conclude it's absolutely convergent. 
But I'm not really sure how I'm supposed to show the series is convergent. I intuitively get it, and I reckon you could do it by some kind of comparison test, but I'm not sure what to compare to. $ \displaystyle \sum_{n=7}^\infty \arctan(1/n)$ diverges, so we can't use that. 
 A: $$0\leq \arctan\left(\frac{1}{n}\right)\leq\frac{1}{n}$$
leads to:
$$ 0\leq \sum_{n\geq 7}\frac{\arctan(1/n)}{n}\leq\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}.$$
A: In order to find an inequality involving $\arctan$, you can use the Mean Value theorem between $0$ and $0 < x < \frac{\pi}{2}$ for the function $$f(x)=\arctan x$$ you have $$\vert \arctan x \vert = \vert (\arctan)^\prime(\xi) \vert \vert x \vert $$ with $\xi \in (0,x)$ and $$\vert (\arctan)^\prime(\xi) \vert = \frac{1}{1+\xi^2}< 1$$ Hence $$0 \le \arctan x \le x$$ for $0<x<\frac{\pi}{2}$.
Then you're done as $0 \le \frac{\arctan\frac{1}{n}}{n} \le \frac{1}{n^2}$ and the series $\sum \frac{1}{n^2}$ converges.
A: As we have a series with positive terms, we can use equivalents:
$$\arctan\frac1n\sim_\infty\frac1n,\quad\text{hence}\quad \frac{\arctan\frac1n}{n}\sim_\infty \frac1{n^2},$$
which converges. Hence the given series converges.
A: $$ x < \tan x\quad(x>0)$$
$$\tan^{-1}x < x$$
$$\sum_{n=1}^{\infty} \frac{\tan^{-1}\frac1n}{n}<\sum_{n=1}^{\infty} \frac{\frac1n}{n}=\frac{\pi^2}6$$
