MHB Sum of Infinite Series: cot^-1(5/sqrt(3))+cot^-1(9/sqrt(3))+...

[Saitama]

Find the sum of the following series upto infinite terms:

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

 

[chisigma]

Find the sum of the following series upto infinite terms:

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

Spoiler

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

$\chi$ $\sigma$

 

[Saitama]

Spoiler

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). :)

 

[chisigma]

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

$\chi$ $\sigma$

 

[Saitama]

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)