Summation of $\arcsin $ series.

Question

What is $a $ if

$$\sum _{n=1} ^{\infty} \arcsin \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n (n+1)}}\right) =\frac {\pi }{a} \,?$$

Attempt: What I tried is to convert the series to $\arctan$ and then convert it telescoping series. So in terms of $\arctan $ it becomes

$$\arctan \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n}+\sqrt {n-1}}\right) $$

but now if I divide by $n$ it simplifies as $n\frac {\pi}4-\sum _1^{\infty} \arctan \left(\frac {\sqrt {n-1}}{\sqrt {n}}\right) $ but as $n$ is tending towards infinity it will lead to infinity which seems wrong. Also note that $a$ is an integer . Thanks!

Answer 1

We can utilize the following trigonometric identity $$\arcsin x-\arcsin y = \arcsin(x\sqrt{1-y^2}-y\sqrt{1-x^2})$$ by putting $x=\frac{1}{\sqrt{n}}$, $y=\frac{1}{\sqrt{n+1}}$. Then we get \begin{align}\arcsin \frac{1}{\sqrt{n}}-\arcsin \frac{1}{\sqrt{n+1}} &= \arcsin\left(\frac{1}{\sqrt{n}}\sqrt{1-\frac{1}{n+1}}-\frac{1}{\sqrt{n+1}}\sqrt{1-\frac{1}{n}}\right)\\ &=\arcsin\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right). \end{align} So we have a telescoping series \begin{align} \sum_{n=1}^{k} \arcsin\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right) &= \sum_{n=1}^{k} \left(\arcsin \frac{1}{\sqrt{n}}-\arcsin \frac{1}{\sqrt{n+1}}\right)\\ &=\arcsin 1-\arcsin \frac{1}{\sqrt{k+1}} \end{align} and in the limit $\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right)=\arcsin 1-\arcsin 0=\frac{\pi}{2}$ and $a=2$.

Answer 2

Taking the principal branch of $\arcsin$ (with values in $\bigl[-\frac{\pi}{2}, \frac{\pi}{2}\bigr]$), we have

$$\tan\bigl(\arcsin s\bigr) = \frac{\sin \bigl(\arcsin s\bigr)}{\cos \bigl(\arcsin s\bigr)} = \frac{s}{\sqrt{1 - s^2}}.$$

With $s = \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}$, we get

\begin{align} 1 - s^2 &= 1 - \frac{(\sqrt{n} - \sqrt{n-1})^2}{n(n+1)}\\ &= \frac{n^2 + n - (n - 2\sqrt{n(n-1)} + n-1)}{n(n+1)}\\ &= \frac{n(n-1) + 2\sqrt{n(n-1)} + 1}{n(n+1)}\\ &= \frac{(1 + \sqrt{n(n-1)})^2}{n(n+1)}, \end{align}

and so

\begin{align} \tan \biggl(\arcsin \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\biggr) &= \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\cdot \frac{\sqrt{n(n+1)}}{1 + \sqrt{n(n-1)}} \\ &= \frac{\sqrt{n} - \sqrt{n-1}}{1 + \sqrt{n} \sqrt{n-1}} \\ &= \tan\bigl(\arctan \sqrt{n} - \arctan \sqrt{n-1}\bigr), \end{align}

whence we obtain

$$\sum_{n = 1}^{\infty} \arcsin \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}} = \sum_{n = 1}^\infty \bigl( \arctan \sqrt{n} - \arctan \sqrt{n-1}\bigr) = \frac{\pi}{2}.$$