Inverse tan series

Geometry Level 3

$\tan \left [ \tan^{-1}\left(\frac12\right)+\tan^{-1}\left(\frac29\right)+\tan^{-1}\left(\frac18\right)+\tan^{-1}\left(\frac2{25}\right)+\tan^{-1}\left(\frac1{18}\right)+\ldots \right]$

What is the value of the expression above?

The answer is 3.

3 solutions

Tanishq Varshney
Jun 22, 2015

$\large{T_{r}=\arctan (\frac{\sqrt{2}}{r+1})^{2}}$

$\large{S=\displaystyle \sum_{r=1}^{\infty} \arctan (\frac{2}{r^2+2r+1})}$

$\large{S=\displaystyle \sum_{r=1}^{\infty} \arctan (\frac{(r+2)-r}{1+r(r+2)})}$

$\large{S=\displaystyle \sum_{r=1}^{\infty} \arctan (r+2)-\arctan (r)}$

$\huge{S=\displaystyle \lim_{n\to \infty} \arctan (\frac{3n^2+7n}{n^2+9n+10})}$

$\large{S=\arctan (3)}$

Very nice problem.

Panya Chunnanonda - 5 years, 11 months ago

Just a modification of the problem asked in JEE 2013.

Satyajit Mohanty - 5 years, 11 months ago

I'm not seeing that manipulation from the third to fourth line. Could you explain please?

Sal Gard - 5 years, 2 months ago

Please show how you got your general term....🙋

Dhruv Joshi - 4 years, 2 months ago

Wei Xian Lim
Jun 23, 2015

Let $U=\sum_{i=1}^n\arctan\left(\frac{1}{2i^2}\right)$ and $T=\sum_{i=1}^n\arctan\left(\frac{2}{(2i+1)^2}\right)$ ,

It can be shown by induction that $\tan U=\frac{n}{n+1}$ and $\tan T=\frac{2n}{4n+5}$ .

Since $S = U + T$ ,

$\tan S = \frac{\tan U+\tan T}{1-\tan U\tan T} = \lim_{n\to\infty}\frac{6n^2+7n}{2n^2+9n+5}=\boxed{3}$

You and a another solution sender have different final term in which you apply limit ....why?

Dhruv Joshi - 4 years, 2 months ago

Jozofrend Horvath
Aug 17, 2016

Run a sequence in QB

Inverse tan series

The answer is 3.

3 solutions

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