# arccot limit: $\sum_{r=1}^{\infty}\cot ^{-1}(r^2+\frac{3}{4})$

I have to find the limit of this sum: $$\sum_{r=1}^{\infty}\cot ^{-1}(r^2+\frac{3}{4})$$

I tried using sandwich theorem , observing:

$$\cot ^{-1}(r^3)\leq\cot ^{-1}(r^2+\frac{3}{4})\leq\cot ^{-1}(r^2)$$

Now when I was calculating the limit of left hand expression, I convert it to $\tan^{-1}$, by using: $$\tan^{-1}\frac{1}{x} = \cot^{-1}x$$

but couldn't sum up the terms of arctan series.

• How can I proceed?

• Is there any better way ?

Hint

The general term can be written as $$\tan^{-1}\frac{1}{r^2+3/4}$$ $$=\tan^{-1}\frac{r+1/2-(r-1/2)}{(r-1/2)(r+1/2)+1}$$ $$=\tan^{-1}(r+1/2)-\tan^{-1}(r-1/2)$$

• Nice! What made you think that way? Thanks! May 21, 2016 at 14:16
• Also can we apply the sandwich theorem some way here? May 21, 2016 at 14:16
• @MaxPayne I doubt it.. May 21, 2016 at 14:18
• He's trying to get you to see that it is a collapsing series. May 21, 2016 at 14:19
• @MaxPayne, $$r^2+\dfrac34=1+r^2-\left(\dfrac12\right)^2$$ May 22, 2016 at 17:05

One may use $$\arctan a - \arctan b=\arctan \left(\frac{a-b}{1+ab} \right), \quad a,b \in \left[0,\frac\pi2\right],$$ with $$a=\frac{2n-3}{4(n+1)},\quad b=\frac{2(n-1)-3}{4n}, \quad n=1,2,3,\ldots,$$ giving, for $n \geq1$, $$\arctan \left(\frac{2n-3}{4(n+1)} \right)-\arctan \left(\frac{2(n-1)-3}{4n} \right)=\arctan \left(\frac1{n^2+\frac34} \right)$$ then, by telescoping, $$\sum_{n=1}^N\arctan \left(\frac1{n^2+\frac34} \right)=\arctan \left(\frac{2N-3}{4(N+1)} \right)-\arctan \left(-\frac34 \right).$$ Letting $N \to \infty$ gives

$$\sum_{n=1}^\infty\arctan \left(\frac1{n^2+\frac34} \right)=\arctan \left(\frac12 \right)+\arctan \left(\frac34 \right)=\arctan 2.$$