Sum of the series cot⁻¹(5/√3) + cot⁻¹(9/√3) + cot⁻¹(15/√3) +

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Saitama
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Find the sum of the following series upto infinite terms:

$$\cot^{-1}\left(\frac{5}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{9}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{15}{\sqrt{3}}\right)+\cot^{-1}\left(\frac{23}{\sqrt{3}}\right)+\cdots$$
 
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Pranav said:
Find the sum of the following series upto infinite terms:

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

Since I'm more familiar with the function $\displaystyle \tan^{-1} x$ let me write the series as...

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

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

$\chi$ $\sigma$
 
chisigma said:
Since I'm more familiar with the function $\displaystyle \tan^{-1} x$ let me write the series as...

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

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

$\chi$ $\sigma$

Hi chisigma!

Thanks for participating, your answer is correct! I am interested in your proof for (2). :)
 
Pranav said:
Hi chisigma!

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

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

Kind regards

$\chi$ $\sigma$
 
chisigma said:
May be that the best for me is to open a math note dedicated to the series of inverse functions...

Kind regards

$\chi$ $\sigma$

Definitely! (Yes)