MHB Evaluation of Infinite sum of Inverse Trig. Series.

[juantheron]

How can we prove $$\displaystyle \tan^{-1}\left(\frac{4}{7}\right)+\tan^{-1}\left(\frac{4}{19}\right)+\tan^{-1}\left(\frac{4}{39}\right)+\tan^{-1}\left(\frac{4}{67}\right)+...\infty = \frac{\pi}{4}+\cot^{-1}(3)$$

My Trial: First we will calculate $\bf{n^{th}}$ terms of Given Series, $$7,19,39,76$$

I have got $4n^2-8n+7$

So we can Write the given series as $$\displaystyle \lim_{n\rightarrow \infty}\sum_{r=1}^{n}\tan^{-1}\left(\frac{4}{4n^2-8n+7}\right)$$

But I did not understand how can i solve it

Help me

Thanks

 

[Saitama]

How can we prove $$\displaystyle \tan^{-1}\left(\frac{4}{7}\right)+\tan^{-1}\left(\frac{4}{19}\right)+\tan^{-1}\left(\frac{4}{39}\right)+\tan^{-1}\left(\frac{4}{67}\right)+...\infty = \frac{\pi}{4}+\cot^{-1}(3)$$

My Trial: First we will calculate $\bf{n^{th}}$ terms of Given Series, $$7,19,39,76$$

I have got $4n^2-8n+7$

So we can Write the given series as $$\displaystyle \lim_{n\rightarrow \infty}\sum_{r=1}^{n}\tan^{-1}\left(\frac{4}{4n^2-8n+7}\right)$$

But I did not understand how can i solve it

Help me

Thanks

Your general term isn't correct. $4n^2+3$ fits better.

So,you have to evaluate the following sum:
$$\sum_{r=1}^{\infty} \arctan\left(\frac{4}{4r^2+3}\right)=\sum_{r=1}^{\infty} \arctan\left(\frac{1}{r^2+3/4}\right)=\sum_{r=1}^{\infty} \arctan\left(\frac{(r+1/2)-(r-1/2)}{1+(r+1/2)(r-1/2)}\right)$$
$$=\sum_{r=1}^{\infty} \arctan\left(r+\frac{1}{2}\right)-\arctan\left(r-\frac{1}{2}\right)$$
This is a telescoping series which should be easy to evaluate.