# How to prove the divergence of the following sum: $\sum^{\infty}_{x=1}\cot^{-1}\left[\frac{1}{1+x^2+x}\right]$

I'm having difficulty proving the divergence of the following infinite sum :

[Note: Here, $\cot^{-1}$ is inverse $\cot$, i.e. arccot]

$$\sum^{\infty}_{x=1}\cot^{-1}\left[\frac{1}{1+x^2+x}\right]$$

I tried the ratio test as well as the root test, but all give indecisive results. Please help me prove the divergence of the above sum (and please do elaborate, ideally without skipping much steps, if using other limit tests as I'm not familiar with them).

## 2 Answers

Recall that $\def\arccot{\operatorname{arccot}}\arccot 0 = \frac\pi 2$, hence, for $x \to \infty$ $$\arccot \frac 1{1 + x + x^2} \to \arccot 0 = \frac\pi 2 \ne 0$$ So, as the sequence of the terms does not converge to zero, the series cannot converge.

• Thanks. Can't believe I missed this... sigh.. – Kugelblitz Mar 21 '16 at 12:30

HINT:

$$\arccot\dfrac1{1+x+x^2}=\dfrac\pi2-\arctan\dfrac1{1+x+x^2}$$

$$\arctan\dfrac1{1+x+x^2}=\arctan\dfrac{x+1-x}{1+x(1+x)}=\arctan(x+1)-\arctan x$$

which is a Telescoping Series