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