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I am trying to find the infinite sum

$$\sum_{n=1}^\infty \cot^{-1} (n^2 + ( \frac{3}{4})),$$ I tried to get a telescopic series but I couldn't find one.

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  • $\begingroup$ have you tried Wolfram Alpha? $\endgroup$ – Dr. Sonnhard Graubner Sep 22 '17 at 13:33
  • $\begingroup$ Actually I am not interested in the value but approach. Can writing the terms in telescopic form and cancelling out terms be applied? $\endgroup$ – jnyan Sep 22 '17 at 13:35
  • $\begingroup$ i think no that we can use this method here $\endgroup$ – Dr. Sonnhard Graubner Sep 22 '17 at 13:36
  • $\begingroup$ @jnyan There is a nice closed form, so I do wonder. $\endgroup$ – George Coote Sep 22 '17 at 13:39
  • $\begingroup$ i have a formula for the finite sum $\endgroup$ – Dr. Sonnhard Graubner Sep 22 '17 at 13:40
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As $\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$

$$\dfrac{4n^2+3}4=1+\dfrac{4n^2-1}4=1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2$$

Again,$\dfrac{2n+1}2-\dfrac{2n-1}2=1$

So, we can write $\cot^{-1}\left(n^2+\dfrac34\right)=\cot^{-1}\left(\dfrac{1+\dfrac{2n+1}2\cdot\dfrac{2n-1}2}{\dfrac{2n+1}2-\dfrac{2n-1}2}\right)$

$=\cot^{-1}\left(\dfrac{2n-1}2\right)-\cot^{-1}\left(\dfrac{2n+1}2\right)$

See also: Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

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Hint. One may use, for $n\ge1$, $$\tan^{-1}\frac{1}{n^2+\frac34}=\tan^{-1}\frac{\left(n+\frac12\right)-\left(n-\frac12\right)}{1+\left(n+\frac12\right)\left(n-\frac12\right)}=\tan^{-1}\left(n+\frac12\right)-\tan^{-1}\left(n-\frac12\right)$$ then use the link between $\cot^{-1}$ and $\tan^{-1}$.

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