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

Question

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 $$\tan^{-1} x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$ — Qurultay, Dec 15 '18 at 05:33
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? — clathratus, Dec 15 '18 at 05:36
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. — Empy2, Dec 15 '18 at 05:51

score 9 · Accepted Answer · answered Dec 15 '18 at 05:51

9

Hint

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

answered Dec 15 '18 at 05:51

C.S.

5,548

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

1 Answers1