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 $$\tan^{-1}\frac{1}{r^2+3/4}$$ $$=\tan^{-1}\frac{r+1/2-(r-1/2)}{(r-1/2)(r+1/2)+1}$$ $$=\tan^{-1}(r+1/2)-\tan^{-1}(r-1/2)$$

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

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