Interesting Arctangent Sums

Post your own problems!! Hard Sums / Products
#1

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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#2

(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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#3

was that meant to be \dfrac{\pi}{4} or have i gone wrong somewhere?

#4

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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#5

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?)

#6

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

#7

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)

#8

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

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