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sum with arcot

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Post Thu Mar 17, 2016 11:48 am
galactus User avatar
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Posts: 902
Here is a cool sum. Is it possible to derive this closed form?.

I apologize for the wacky sums that did not pan out. As I said, I only had around 10 decimals places. It was safe to assume that closed forms like that were too good to be true :)
They came from that Inverse Symbolic Calculator. I rarely use it, so I was just playing around.


Anyway, try this one if you like. I think it is more amenable.

$$\sum_{n=1}^{\infty}\cot^{-1}(n^{2})=\cot^{-1}\left(\frac{1+z}{1-z}\right)$$

where $z=\cot(\frac{\pi}{\sqrt{2}})\tanh(\frac{\pi}{\sqrt{2}})$

Post Thu Mar 17, 2016 11:39 pm

Posts: 47

galactus wrote:

$$\sum_{n=1}^{\infty}\cot^{-1}(n^{2})=\cot^{-1}\left(\frac{1+z}{1-z}\right)$$

where $z=\cot(\frac{\pi}{\sqrt{2}})\tanh(\frac{\pi}{\sqrt{2}})$


Well, the basic idea is the following:

\begin{align*}
\sum_{n=1}^{\infty}{\rm arccot} n^2 &= \sum_{n=1}^{\infty} \arctan \frac{1}{n^2} \\
&= -\arg \prod_{n=1}^{\infty} \left ( 1- \frac{i}{n^2} \right )\\
&= \left[ -\arg \prod_{n=1}^{\infty} \left ( 1-\frac{z^2}{n^2} \right ) \right]_{z=\frac{\sqrt{2}}{2}+ i \frac{\sqrt{2}}{2}}\\
&= \left[ -\arg \left ( \frac{\sin \pi z}{\pi z} \right ) \right]_{z=\frac{\sqrt{2}}{2}+ i \frac{\sqrt{2}}{2}} \\
&=\frac{\pi}{4} - \arctan \left ( \cot \frac{\pi \sqrt{2}}{2} \tanh \frac{\pi \sqrt{2}}{2} \right )
\end{align*}

which is equivelant to what Galactus posted if we make use of basic trig. identities. Yeah, that series is old chestnut. :)

Post Thu Mar 17, 2016 11:41 pm

Posts: 47

Hey folks, how about these two sums?

$$\sum_{n=1}^{\infty}\arctan \frac{2}{n^2} \quad, \quad \sum_{n=1}^{\infty} {\rm arccot} \frac{n^2}{8}$$

Post Mon Apr 18, 2016 8:44 am

Posts: 47

VJKey wrote:
Hey folks, how about these two sums?

$$\sum_{n=1}^{\infty}\arctan \frac{2}{n^2} \quad, \quad \sum_{n=1}^{\infty} {\rm arccot} \frac{n^2}{8}$$


Let me bump this. Although they're quite easy.

Hint: Telescope both of them.

Post Fri Apr 22, 2016 3:06 am

Posts: 27
VJKey wrote:
Hey folks, how about these two sums?

$$\sum_{n=1}^{\infty}\arctan \frac{2}{n^2} \quad, \quad \sum_{n=1}^{\infty} {\rm arccot} \frac{n^2}{8}$$


The $n^{\text{th}}$ partial sum of the first series is $\tan^{-1}(n+1)+\tan^{-1}{n} - \tan^{-1}{1}$

Take the limit to infinity of $n$, and the sum is $\frac{3\pi}{4}$

The $n^{\text{th}}$ partial sum of the second series is $\cot^{-1}{0} + \cot^{-1}{1} - \cot^{-1}{\frac{n-1}{2}} - \cot^{-1}{\frac{n}{2}} - \cot^{-1}{\frac{n+1}{2}} - \cot^{-1}{\frac{n+2}{2}}$

Take the limit to infinity of $n$, and the sum is $\frac{3\pi}{4}$

Post Wed May 04, 2016 10:14 pm

Posts: 1
you were right.it was good indeed,
Graduated from Soran University with First Class Degree with Honours in Computer Science.


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