Sum of the series $\csc^{-1} \sqrt{10}+ \csc^{-1} \sqrt{50}+\csc^{-1}\sqrt{170}...$

Question

Find the sum of the series: $\csc^{-1}\sqrt{10}+ \csc^{-1}\sqrt{50}+\csc^{-1}\sqrt{170}...\csc^{-1}\sqrt{(n^2+1)(n^2+2n+2)}$

I converted the series to $\sum^{n} _{i=0}\arcsin \dfrac{1}{\sqrt{(i^2+1)(i^2+2i+2)}}$ and then tried to use $\arcsin(x)+\arcsin(y)= \arcsin({x\sqrt{1+y^2}+y\sqrt{1+x^2}}) \forall x,y \ge 0 $ and $x^2+y^2\le1$.

There was no clear pattern in the terms obtained. How to solve this question? And what is the general trick to solve inverse trigonometric summations?

Answer 1

The given sum can be represented as $$\tan ^{-1} \frac {1}{3}+\tan ^{-1} \frac {1}{7}+\tan ^{-1} \frac {1}{13}\cdots \tan ^{-1} \frac {1}{n^2+n+1}$$ This can be represented as $$\sum_{k=1}^n \tan ^{-1} \left(\frac {1}{k^2+k+1}\right)$$

$$\sum_{k=1}^n \tan ^{-1} \left(\frac {(k+1)-(k) }{1+(k+1)(k)}\right) $$ $$\sum_{k=1}^n \tan ^{-1} (k+1) -\tan ^{-1} (k)$$

Which telescopes to $$\tan^{-1} (n+1)- \frac {\pi}{4}$$

Answer 2

Use $$\arcsin\frac{1}{\sqrt{n^2+1}}-\arcsin\frac{1}{\sqrt{n^2+2n+2}}=\arcsin\frac{1}{\sqrt{(n^2+1)(n^2+2n+2)}}$$ and the telescopic sum.

Indeed, $$\sin\left(\arcsin\frac{1}{\sqrt{n^2+1}}-\arcsin\frac{1}{\sqrt{n^2+2n+2}}\right)=$$ $$=\frac{1}{\sqrt{n^2+1}}\cdot\frac{n+1}{\sqrt{n^2+2n+2}}-\frac{n}{\sqrt{n^2+1}}\cdot\frac{1}{\sqrt{n^2+2n+2}}=\frac{1}{\sqrt{(n^2+1)(n^2+2n+2)}}.$$

Id est, $$\sum_{i=1}^n\arcsin\frac{1}{\sqrt{(i^2+1)(i^2+2i+2)}}=\sum_{i=1}^n\left(\arcsin\frac{1}{\sqrt{i^2+1}}-\arcsin\frac{1}{\sqrt{i^2+2i+2}}\right)=$$ $$=\arcsin\frac{1}{\sqrt{2}}-\arcsin\frac{1}{\sqrt{n^2+2n+2}}=\frac{\pi}{4}-\arcsin\frac{1}{\sqrt{n^2+2n+2}}.$$

The Manthanein's answer and mine they are the same.

Indeed, we need to prove that $$\arcsin\frac{1}{\sqrt{n^2+2n+2}}+\arctan(n+1)=\frac{\pi}{2}$$ or $$\cos\left(\arcsin\frac{1}{\sqrt{n^2+2n+2}}+\arctan(n+1)\right)=0$$ or $$\sqrt{1-\frac{1}{n^2+2n+2}}\cdot\frac{1}{\sqrt{1+(n+1)^2}}-\frac{1}{\sqrt{n^2+2n+2}}\cdot\sqrt{1-\frac{1}{1+(n+1)^2}}=0,$$ which is obvious.