This question already has an answer here:

$$\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{4}{4r^2 +3}\right)\right)= ? $$

I wrote it in the form:

$$\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{\dfrac43}{\dfrac{4r^2}{3} +1}\right)\right)$$ and tried to use: $$\arctan x- \arctan y = \arctan\left(\dfrac{x-y}{1+xy}\right)$$ but that trick doesn't help here. How to go about solving this problem then?


marked as duplicate by lab bhattacharjee calculus May 21 '18 at 18:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


You are on the right track. Just note that $$\frac{4}{4r^2+3} =\frac{1}{r^2+3/4}=\frac{1}{1+r^2-(1/2)^2}=\frac{(r+1/2)-(r-1/2)}{1+(r+1/2)(r-1/2)}$$ and therefore $$\arctan\left(\frac{4}{4r^2+3}\right)=\arctan\left(r+\frac{1}{2}\right)- \arctan\left(r-\frac{1}{2}\right)\\=\arctan\left((r+1)-\frac{1}{2}\right)- \arctan\left(r-\frac{1}{2}\right).$$ Can you take it from here?

  • $\begingroup$ @Abcd Have you found the final answer? $\endgroup$ – Robert Z May 21 '18 at 15:01
  • $\begingroup$ Thanks for the help! $\endgroup$ – Abcd May 21 '18 at 18:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.