# Summation of arctangents

#### Pranav

##### Well-known member
Problem:
Evaluate
$$\lim_{n\rightarrow \infty} \left(\sum_{r=1}^n (\arctan(2r^2))-\frac{n\pi}{2}\right)$$

Attempt:
I tried evaluating the summation but couldn't. Had the problem involved $\arctan(1/(2r^2))$, I could rewrite it as
$$\arctan\left(\frac{2r+1-(2r-1)}{1+(2r+1)(2r-1)}\right)$$
and evaluating the sum would be quite easy but honestly, I have no idea for the given problem.

Any help is appreciated. Thanks!

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I tried evaluating the summation but couldn't. Had the problem involved $\arctan(1/(2r^2))$, I could rewrite it as
$$\arctan\left(\frac{2r+1-(2r-1)}{1+(2r+1)(2r-1)}\right)$$
and evaluating the sum would be quite easy but honestly, I have no idea for the given problem.

Any help is appreciated. Thanks!
Maybe you could use that $$\displaystyle \arctan (x) + \arctan \left( \frac{1}{x}\right) =\frac{\pi}{2}$$ for $$\displaystyle x>0$$.

#### chisigma

##### Well-known member
Problem:
Evaluate
$$\lim_{n\rightarrow \infty} \left(\sum_{r=1}^n (\arctan(2r^2))-\frac{n\pi}{2}\right)$$

Attempt:
I tried evaluating the summation but couldn't. Had the problem involved $\arctan(1/(2r^2))$, I could rewrite it as
$$\arctan\left(\frac{2r+1-(2r-1)}{1+(2r+1)(2r-1)}\right)$$
and evaluating the sum would be quite easy but honestly, I have no idea for the given problem.

Any help is appreciated. Thanks!
Using the identity...

$\displaystyle \sum_{k=1}^{n} \tan^{-1} (2\ k^{2}) - n\ \frac{\pi}{2} = - \sum_{k=1}^{n} \tan^{-1} \frac{1}{2\ k^{2}}\ (1)$

... the series becomes $\displaystyle - \sum_{k=1}^{\infty} \tan^{-1} \frac{1}{2\ k^{2}}$. Following the procedure described in...

http://mathhelpboards.com/math-notes-49/series-inverse-functions-8530.html

... we have $\displaystyle G(u,v) = \frac{u - v}{1 + u\ v}$ and choosing the sequence $c_{k}= \frac{k}{k+1}$ which is stricktly increasing and tends to 1, we obtain $\displaystyle G(c_{k},c_{k-1}) = \frac{1}{2\ k^{2}}$ and finally...

$\displaystyle \sum_{k=1}^{\infty} \tan^{-1} \frac{1}{2\ k^{2}} = \tan^{-1} 1 - \tan^{-1} 0 = \frac{\pi}{4}\ (2)$

Kind regards

$\chi$ $\sigma$

Last edited:

#### Pranav

##### Well-known member
Using the identity...

$\displaystyle \sum_{k=1}^{n} \tan^{-1} (2\ k^{2}) - n\ \frac{\pi}{2} = \sum_{k=1}^{n} \tan^{-1} \frac{1}{2\ k^{2}}\ (1)$
Ah, that was quite silly on my part, I was so much involved in evaluating the summation, I did not think of the above identity, thanks a lot chisigma!

BTW, you have a sign error.
... the series becomes $\displaystyle \sum_{k=1}^{\infty} \tan^{-1} \frac{1}{2\ k^{2}}$. Following the procedure described in...
That's the same series I mentioned in my attempt, it's easy to evaluate this.