# Sum of $50$ terms of $\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21+.....$ [duplicate]

Find the sum of the first $$50$$ terms of the series $$\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21+.....$$

$$\sum_1^{50}=\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21+.....\\ =\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}+.....=$$ My reference gives the solution $$\tan^{-1}\dfrac{5}{6}$$, but I do not have any clue of doing it ?

Note: I know that $$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$$ if $$xy<1$$.

• Note that $$\tan^{-1} x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$ Dec 15, 2018 at 5:33
• what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $\sum$ notation? Dec 15, 2018 at 5:36
• @clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$. Dec 15, 2018 at 5:41
• $a_n=1+n+n^2$, I guess Dec 15, 2018 at 5:45
• Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it. Dec 15, 2018 at 5:51

• Use the fact that $$\cot^{-1}(x)=\tan^{-1}\frac{1}{x}$$
• $$\tan^{-1}\left(\frac{1}{n^2+n+1}\right) = \tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) =\tan^{-1}(n+1) -\tan^{-1}(n)$$