Interesting Arctangent Sums

These two sums are basically unconnected but look similar so I’ll lump them in one post.

(i)

Prove that for x > y > 0:

\displaystyle \arctan x - \arctan y = \arctan \left(\frac {x - y} {1 + xy}\right)

Deduce:

\displaystyle \sum_{n = 1}^\infty \arctan \left(\frac 2 {n^2}\right) = \frac {3\pi} 4

(ii) Show that:

\displaystyle \sum_{n = 1}^\infty \arctan \left(\frac 1 {8n^2}\right) = \frac \pi 4 - \arctan \left(\tanh \frac \pi 4\right)

You may assume that:

\displaystyle \sinh x = x \prod_{n = 1}^\infty \left(1 + \frac {x^2} {n^2 \pi^2}\right)

for all x \in \mathbb C without proof.

[(ii) originally posted by me on TSR, (i) classic exercise and much easier]

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(i)
we know that

\begin{align} \arctan x - \arctan y &= \arctan\left(\tan\left(\arctan x - \arctan y\right)\right) \\ &= \arctan\left(\frac{\tan(\arctan x)-\tan(\arctan y)}{1+\tan(\arctan x)\tan(\arctan y)}\right) \\ &= \arctan\left(\frac{x-y}{1+xy}\right) \end{align}

now for the sum, consider letting xy=n^2-1 and x-y=2. we must also acknowledge that n\geqslant 1 so that n^2-1 \geqslant 0\iff xy+1 \geqslant 1 (so denominator is never 0 this way) and that x=y+2>y.

\begin{align} \implies y(y+2)&=n^2-1 \\ \iff y^2 +2y+(1-n^2)&=0 \\ \iff y&=\frac{-2±\sqrt{2^2-4(1-n^2)}}{2} \\ &= -1±\frac{2\sqrt{1-1+n^2}}{2} \\ &= -1 ± n \end{align}

but n\geqslant 1 and y > 0 so we can conclude that y=n-1, x=n+1.

\begin{align} \therefore \arctan\left(\frac{2}{1+(n^2-1)}\right) &= \arctan(n+1) - \arctan(n-1) \\ \implies \sum_{n=1}^{\infty} \arctan\left(\frac{2}{1+(n^2-1)}\right)&= \sum_{n=1}^{\infty} (\arctan(n+1) - \arctan(n-1)) \\ \iff \sum_{n=1}^{\infty} \arctan\left(\frac{2}{n^2}\right) &=\sum_{n=1}^{\infty} \arctan(n+1) - \sum_{n=1}^{\infty} \arctan(n-1) \\ &= \lim_{N\to\infty}\sum_{n=1}^{N} \arctan(n+1) - \lim_{N\to\infty}\sum_{n=1}^{N} \arctan(n-1) \\ &= \lim_{N\to\infty}\sum_{n=3}^{N+2} \arctan(n-1) - \lim_{N\to\infty}\sum_{n=1}^{N} \arctan(n-1) \\ &= \lim_{N\to\infty}\left(\arctan(N+1)+\arctan N-\arctan1-\arctan0\right) \\ &= \lim_{N\to\infty}\left(2\arctan N-\frac \pi 4-0\right) \\ &= \frac {2\pi}{2} -\frac \pi 4 \\ &= \frac{3\pi}{4} \end{align}
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was that meant to be \dfrac{\pi}{4} or have i gone wrong somewhere?

I meant to write \displaystyle \arctan \left(\frac 1 {2n^2}\right)!

Your result is correct for \displaystyle \arctan \left(\frac 2 {n^2}\right) so I’ll change the question accordingly.

Turns out I’d made the same typo on the second one too, oops.

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ohh lmao, worked out nicely tho. you could add the other one as a part 2 then finish with the last one (does last one require any undergrad knowledge?)

It requires knowledge of complex logarithms, using the method I know, so yeah I’d say so.

k, i’ll just have to leave it for another day then
are they related to complex exponentials? (just due to exponential and natural log being inverses)

Yeah - if you want to read more about it: https://en.wikipedia.org/wiki/Complex_logarithm

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