How to prove $\sum_{n=1}^\infty\operatorname{arccot}\frac{\sqrt[2^n]2+\cos\frac\pi{2^n}}{\sin\frac\pi{2^n}}=\operatorname{arccot}\frac{\ln2}\pi$? [closed]

Solution 1:

Rewrite the sum as

$$\sum_{n=1}^{\infty} \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}+\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$

Now let

$$a_n = \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$

Then one may show (nontrivial!) that

$$a_n - a_{n-1} = \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}+\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$

so that we have a telescoping sum:

$$\sum_{n=1}^{\infty} (a_n-a_{n-1}) = a_{\infty}-a_0$$

where

$$a_{\infty} = \lim_{n\to\infty} a_n $$

Now,

$$\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}} \sim \frac{\pi/2^n}{1+\frac{\log{2}}{2^n}-1} = \frac{\pi}{\log{2}}$$

Note also that $a_0=0$. therefore the sum is simply

$$\arctan{\frac{\pi}{\log{2}}}$$

which is the stated result.

ADDENDUM

For completeness, I'll outline a few steps to demonstrate how to prove the difference equation above. Start with the relation

$$\arctan{p}-\arctan{q} = \arctan{\frac{p-q}{1+p q}}$$

so we need to work with the following argument of the arctangent:

$$\frac{\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}} - \frac{\sin{\left (\frac{2\pi}{2^n}\right)}}{2^{2^{1-n}}-\cos{\left (\frac{2\pi}{2^n}\right)}}}{1+\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}}\frac{\sin{\left (\frac{2\pi}{2^n}\right)}}{2^{2^{1-n}}-\cos{\left (\frac{2\pi}{2^n}\right)}}}$$

which simplifies somewhat to

$$\frac{-2^{2^{-n}} \sin \left(\pi 2^{1-n}\right)+2^{2^{1-n}} \sin \left(\pi 2^{-n}\right)+\sin \left(\pi 2^{1-n}\right) \cos \left(\pi 2^{-n}\right)-\sin \left(\pi 2^{-n}\right) \cos \left(\pi 2^{1-n}\right)}{2^{3\ 2^{-n}}+\sin \left(\pi 2^{1-n}\right) \sin \left(\pi 2^{-n}\right)-2^{2^{-n}} \cos \left(\pi 2^{1-n}\right)+\cos \left(\pi 2^{-n}\right) \cos \left(\pi 2^{1-n}\right)-2^{2^{1-n}} \cos \left(\pi 2^{-n}\right)}$$

Now you may show that the numerator is equal to

$$\sin \left(\pi 2^{-n}\right) \left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi 2^{-n}\right)+1\right)$$

and the denominator is equal to

$$2^{2^{-n}} \left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi 2^{-n}\right)+1\right)+\cos \left(\pi 2^{-n}\right) \left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi 2^{-n}\right)+1\right)$$

Cancelling the common factors in the numerator and denominator produces the desired result.