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I am trying to evaluate the infinite sum:

$$\ \sum_{n=0}^{\infty} \sin^{-1} \left( \frac{2 (n^2 + n + 1)}{(n^2 + 1) (n^2 + 2n + 2)} \right) \ = \pi $$

I suspect that the terms inside the arcsin function might allow for a telescoping sum, but I am unsure how to properly manipulate the expression to confirm this. If the terms do telescope, I would like to see a detailed derivation of how the cancellation works. Additionally, if there are alternative approaches (such as using trigonometric identities or series transformations), I would love to explore those as well. Any insights or step-by-step guidance would be greatly appreciated!

Uvindu Dassanayaka
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2 Answers2

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Let $\sin\theta=x$, so $\cos\theta=\sqrt{(1-x^2)}$

$\therefore \tan\theta=\frac{x}{\sqrt{1-x^2}}$

From here we get $\sin^{-1}{(x)}=\tan^{-1}(\frac{x}{\sqrt{1-x^2}})$

Here, $x=\frac{2(n^2+n+1)}{(n^2+1)(n^2+2n+2)}$

So, $\sqrt{1-x^2}=\frac{(n^2+n)(n^2+n+2)}{(n^2+1)(n^2+2n+2)}$

Also using $\tan^{-1}(\frac{2x}{1-x^2})=2\tan^{-1}(x)$

our sum is $S=\sum_{n=0}^{n=\infty}2\tan^{-1}(\frac{1}{n^2+n+1})=\tan^{-1}(\frac{(n+1)-(n)}{1+(n)(n+1)})$

Using $\tan^{-1}(x)-\tan^{-1}(y)=\tan^{-1}(\frac{x-y}{1+xy})$

Now $S=2\sum_{n=0}^{n=\infty} \tan^{-1}(n+1)-\tan^{-1}(n)=\lim_{n\to\infty}[\tan^{-1}({n+1})-\tan^{-1}(0)]$

$\boxed{\therefore S=\pi}$

CrSb0001
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Nucleo
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    On line 5, a $)$ is missing, it should be $(n^2 + n +2)$ – user25406 Feb 04 '25 at 11:41
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    @user25406 I've fixed it. – CrSb0001 Feb 04 '25 at 20:06
  • Once you get to the inverse tangent sum, there is a very nice geometric interpretation. Let $O$ be the origin, let $A_n$ be the point $(1,n)$ for $n\ge0$ and let $\theta_n$ be the angle $A_nOA_{n+1}$. Then you can confirm that $\theta_n=\arctan(1/(n^2+n+1))$, and it is "visually obvious" that $\sum\theta_n=\pi/2$. – David Feb 05 '25 at 00:53
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Hint: if we denote \begin{equation*} x_n=\dfrac{2(n^2+n+1)}{(n^2+1)(n^2+2n+2)},~\text{then}~\sqrt{1-x_n^2}=\frac{(n^2+n)(n^2+n+2)}{(n^2+1)(n^2+2n+2)}. \end{equation*} Deduce that \begin{equation*} \arcsin x_n=\arctan\left(\frac{2(n^2+n+1)}{(n^2+n)(n^2+n+2)}\right)=2\arctan\left(\frac{1}{n^2+n+1}\right)=2\left(\arctan\frac{1}{n}-\arctan\frac{1}{n+1}\right) \end{equation*} and conclude.

Remark. In the formulas above we rely on the trigonometric identity $\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$.

richrow
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