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I have to find the limit of this sum: $$\sum_{r=1}^{\infty}\cot ^{-1}(r^2+\frac{3}{4})$$

I tried using sandwich theorem , observing:

$$\cot ^{-1}(r^3)\leq\cot ^{-1}(r^2+\frac{3}{4})\leq\cot ^{-1}(r^2)$$

Now when I was calculating the limit of left hand expression, I convert it to $\tan^{-1}$, by using: $$\tan^{-1}\frac{1}{x} = \cot^{-1}x$$

but couldn't sum up the terms of arctan series.

  • How can I proceed?

  • Is there any better way ?

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2 Answers 2

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Hint

The general term can be written as $$\tan^{-1}\frac{1}{r^2+3/4}$$ $$=\tan^{-1}\frac{r+1/2-(r-1/2)}{(r-1/2)(r+1/2)+1}$$ $$=\tan^{-1}(r+1/2)-\tan^{-1}(r-1/2)$$

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    $\begingroup$ Nice! What made you think that way? Thanks! $\endgroup$
    – Max Payne
    May 21, 2016 at 14:16
  • $\begingroup$ Also can we apply the sandwich theorem some way here? $\endgroup$
    – Max Payne
    May 21, 2016 at 14:16
  • $\begingroup$ @MaxPayne I doubt it.. $\endgroup$
    – Nikunj
    May 21, 2016 at 14:18
  • $\begingroup$ He's trying to get you to see that it is a collapsing series. $\endgroup$ May 21, 2016 at 14:19
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    $\begingroup$ @MaxPayne, $$r^2+\dfrac34=1+r^2-\left(\dfrac12\right)^2$$ $\endgroup$ May 22, 2016 at 17:05
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One may use $$ \arctan a - \arctan b=\arctan \left(\frac{a-b}{1+ab} \right), \quad a,b \in \left[0,\frac\pi2\right], $$ with $$ a=\frac{2n-3}{4(n+1)},\quad b=\frac{2(n-1)-3}{4n}, \quad n=1,2,3,\ldots, $$ giving, for $n \geq1$, $$ \arctan \left(\frac{2n-3}{4(n+1)} \right)-\arctan \left(\frac{2(n-1)-3}{4n} \right)=\arctan \left(\frac1{n^2+\frac34} \right) $$ then, by telescoping, $$ \sum_{n=1}^N\arctan \left(\frac1{n^2+\frac34} \right)=\arctan \left(\frac{2N-3}{4(N+1)} \right)-\arctan \left(-\frac34 \right). $$ Letting $N \to \infty$ gives

$$ \sum_{n=1}^\infty\arctan \left(\frac1{n^2+\frac34} \right)=\arctan \left(\frac12 \right)+\arctan \left(\frac34 \right)=\arctan 2. $$

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