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#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

$$\cot^{-1}\left(\frac{5}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{9}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{15}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{23}{\sqrt{3}}\right)+\cdots$$

- Thread starter Pranav
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- Thread starter
- #1

- Nov 4, 2013

- 428

$$\cot^{-1}\left(\frac{5}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{9}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{15}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{23}{\sqrt{3}}\right)+\cdots$$

- Feb 13, 2012

- 1,704

$$\cot^{-1}\left(\frac{5}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{9}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{15}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{23}{\sqrt{3}}\right)+\cdots$$

$\displaystyle S = \sum_{n=1}^{\infty} \tan^{-1} \frac{\sqrt{3}}{n^{2} + n + 3}\ (1)$

We can use the general formula...

$\displaystyle \sum_{n=1}^{\infty} \tan^{-1} \frac{c}{n^{2} + n + c^{2}} = \tan^{- 1} c\ (2)$

... obtaining...

$\displaystyle S = \tan^{-1} \sqrt{3} = \frac{\pi}{3}\ (3)$

The prove of (2) will be supplied in a successive post...

Kind regards

$\chi$ $\sigma$

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- #3

- Nov 4, 2013

- 428

Hi chisigma!

$\displaystyle S = \sum_{n=1}^{\infty} \tan^{-1} \frac{\sqrt{3}}{n^{2} + n + 3}\ (1)$

We can use the general formula...

$\displaystyle \sum_{n=1}^{\infty} \tan^{-1} \frac{c}{n^{2} + n + c^{2}} = \tan^{- 1} c\ (2)$

... obtaining...

$\displaystyle S = \tan^{-1} \sqrt{3} = \frac{\pi}{3}\ (3)$

The prove of (2) will be supplied in a successive post...

Kind regards

$\chi$ $\sigma$

Thanks for participating, your answer is correct! I am interested in your proof for (2).

- Feb 13, 2012

- 1,704

May be that the best for me is to open a math note dedicated to the series of inverse functions...Hi chisigma!

Thanks for participating, your answer is correct! I am interested in your proof for (2).

Kind regards

$\chi$ $\sigma$

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- #5

- Nov 4, 2013

- 428

Definitely!May be that the best for me is to open a math note dedicated to the series of inverse functions...

Kind regards

$\chi$ $\sigma$