# Find the sum : $\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$

Problem :

Find the sum of :

$$\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$$

My approach :

Here the $n$'th term is given by :

$$t_n = \sin^{-1}\left[\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right]$$

From now how to proceed further please suggest thanks....

• I inputted this into Mathematica but it did not give an exact result however it gives 1.5708 as an approximate result. I'm guessing the answer is $\frac{\pi}{2}$. – Cameron Williams Jul 29 '13 at 3:12
• Actually what interests me more is a possible geometric proof of this fact, now that would be cool. – Lord Soth Jul 29 '13 at 3:40

This is not an independent answer but a response to Lord Soth's speculation of a geometric proof.

Consider two right-handed triangles $ABC$ and $ABD$ with base $1$ and heights $\sqrt{n}$ and $\sqrt{n-1}$. Let $\theta$ be the angle $\measuredangle CBD$. The area of the triangle $BCD$ can be computed in two ways:

• $\frac12 |BC||BD| \sin\theta = \frac12 \sqrt{n+1} \sqrt{n} \sin\theta$
• $\frac12 |CD||AB| = \frac12 ( \sqrt{n} - \sqrt{n-1} )$

Equate them gives us:

$$\sin \theta = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}$$

On the other handle,

\begin{align} \theta &= \measuredangle CBD = \measuredangle CBA - \measuredangle DBA\\ &= \sin^{-1}\frac{|AC|}{|BC|} - \sin^{-1}\frac{|AD|}{|BD|} = \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}} \end{align}

We obtain

$$\sin^{-1}\left( \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right) = \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}}$$

and we have turned the original series into a telescoping one. The partial sum of the first $n$ terms of original series becomes the angle $\measuredangle ABC$. When $n \to \infty$, the line $BC$ becomes vertical and this geometrically justify why the limit of the series is $\frac{\pi}{2}$.

• +1 Why is this not an answer? It looks like a perfectly fine answer to me. – Lord Soth Jul 29 '13 at 5:09
• @LordSoth it is a supplement to your answer. okay, I'll change the wordings. – achille hui Jul 29 '13 at 5:14

Hint: Use the fact that $\sin^{-1}a+\sin^{-1}b = \sin^{-1}(a\sqrt{1-b^2}+b\sqrt{1-a^2})$. The first two terms then give you $\sin^{-1}(\sqrt{2/3})$. Then applying the same identity with this term and the third term gives you $\sin^{-1}(\sqrt{3/4})$, and...

Edit: Induction step gives some headache, so let me write it for completeness. Suppose the sum of the first $n$ terms are $a = \sin^{-1}\sqrt\frac{n}{n+1}$. We show that the sum of the first $n+1$ terms are $\sin^{-1}\sqrt{\frac{n+1}{n+2}}$. Let $$b = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}$$ denote the $(n+1)$th term. Then, it is sufficient to prove that $c = a\sqrt{1-b^2}+b\sqrt{1-a^2} = \sqrt\frac{n+1}{n+2}$. Indeed, we have \begin{align} c &= \sqrt\frac{n}{n+1}\sqrt{1-\frac{(\sqrt{n+1}-\sqrt{n})^2}{(n+1)(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}\sqrt{1-\frac{n}{n+1}}\\ & = \sqrt\frac{n}{n+1}\sqrt{\frac{n^2+n+1+2\sqrt{n(n+1)}}{(n+1)(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)\sqrt{(n+2)}} \end{align} Noting that the nested radical $\sqrt{n^2+n+1+2\sqrt{n(n+1)}}$ is equal to $1+\sqrt{n(n+1)}$, we obtain $$c = \sqrt n \frac{1+\sqrt{n(n+1)}}{(n+1)\sqrt{(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}} {(n+1)\sqrt{(n+2)}} = \sqrt\frac{n+1}{n+2}$$.

• Trying to prove the general from for $n$ ends up with a real mess. Did you prove this? – nbubis Jul 29 '13 at 3:32
• @nbubis You should be able to go by induction. Assume the sum of the first $n$ terms are equal to $\sin^{-1}\sqrt{\frac{n}{n+1}}$. Then show that the sum of the first $n+1$ terms are $\sin^{-1}\sqrt{\frac{n+1}{n+2}}$ using the formula above (and the formula for the $n$th term of the sequence that the OP has provided). – Lord Soth Jul 29 '13 at 3:37
• @nbubis I wrote down the details. – Lord Soth Jul 29 '13 at 4:30