MHB Evaluation of Infinite sum of Inverse Trig. Series.

AI Thread Summary
The discussion focuses on proving the infinite sum of the series involving inverse tangent functions, specifically showing that the sum equals \(\frac{\pi}{4} + \cot^{-1}(3)\). The initial attempt involved calculating the \(n^{th}\) term of the series, which was incorrectly identified as \(4n^2 - 8n + 7\). A participant corrected this to \(4n^2 + 3\) and provided a reformulation of the series as a telescoping series. The final suggestion indicates that the telescoping nature of the series simplifies the evaluation process significantly. The conversation emphasizes the importance of correctly identifying the general term for accurate summation.
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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
 
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jacks said:
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.
 
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