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Series Contest (SC2)

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Post Mon Jun 17, 2013 4:41 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Cool, zer98! :) Here is another proof using the method of differences -
\(\displaystyle \;\)
\[\begin{aligned}\sum_{k=1}^N 4^k \sin^4(2^{-k}\theta) &= \sum_{k=1}^N 4^k \sin^2 (2^{-k}\theta) (1-\cos^2(2^{-k}\theta)) \\
&= \sum_{k=1}^N 4^k \sin^2(2^{-k}\theta) - \sum_{k=1}^N 4^k \sin^2(2^{-k}\theta)\cos^2(2^{-k}\theta)\\
&= \sum_{k=1}^N 4^k \sin^2(2^{-k}\theta) - \sum_{k=1}^N 4^{k-1} \sin^2(2^{-k+1}\theta)
\end{aligned}\]

Everything cancels except
\[\sum_{k=1}^N 4^k \sin^4(2^{-k}\theta) = 4^N \sin^2(2^{-N}\theta)-\sin^2(\theta)\]
Passing \(N\to \infty\), we arrive at the required result.

Post Mon Jun 17, 2013 5:56 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

This contest will keep running until 30 series are solved.

Problem 21

Prove that
\[\sum_{n=0}^\infty \frac{(2n+1)\text{tanh} \left( \frac{2n+1}{2}\pi \right)}{(2n+1)^4+1} = \frac{\pi}{8}\frac{\cosh \left( \frac{\pi}{\sqrt{2}}\right)-\cos \left(\frac{\pi}{\sqrt{2}}\right)}{\cosh \left( \frac{\pi}{\sqrt{2}}\right)+\cos \left(\frac{\pi}{\sqrt{2}}\right)}\]

Post Tue Jun 18, 2013 5:20 pm
galactus User avatar
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Posts: 902
Hey E. Did you do this one with residues?.

i.e. I got \(\displaystyle (-1/4+i/4)(\sqrt{2}+i+i)\) as one of the poles of \(\displaystyle (2z+1)^{4}+1\)

This led to a residue of \(\displaystyle \frac{\pi (\cosh(a)-\cos(a))}{8(\cosh(a)+\cos(a))}\)

where \(\displaystyle a=\frac{\pi}{\sqrt{2}}\)

by then using \(\displaystyle \sum_{-\infty}^{\infty}f(n)=-(\text{sum of residues of} \pi\cot(\pi z) \text{at the poles of f(z)})\)

The poles of \(\displaystyle \sin(\pi z)\) are at \(\displaystyle z=k\).

The poles of \(\displaystyle \cosh(\frac{\pi}{2}(2z+1))\) are at \(\displaystyle z=\frac{-1}{2}+\frac{2k+1}{2}i\)

So, the residue for each is \(\displaystyle \sum_{k=0}^{\infty}\frac{(2k+1)\tanh(\frac{\pi}{2}(2k+1))}{(2k+1)^{4}+1}\)

Post Wed Jun 19, 2013 3:11 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Yes, this is how I did it. I have another tough one:

Problem 22
\[\sum_{n=1}^\infty \frac{1}{n(e^{2\pi n}-1)}=-\frac{1}{6} \log \left( \frac{\pi^{\frac{3}{2}}}{8 \Gamma^6 \left( \frac{3}{4}\right)}\right)-\frac{\pi}{12}\]

I think there is some kind of Elliptic Stuff involved in this one.

Post Thu Jun 20, 2013 5:43 pm
galactus User avatar
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Posts: 902
Hey E.

As I mentioned in the other post under Series, this can be done by using the identity

\(\displaystyle \prod_{k=1}^{\infty}(1-e^{2k\pi})=\frac{{\pi}^{1/4}e^{\frac{\pi}{12}}}{\sqrt{2}\Gamma(3/4)}\)

This is one of your Theta functions, E :)

\(\displaystyle \frac{e^{-2\pi k}}{n(1-e^{-2\pi k})}=\sum_{n=1}^{\infty}\frac{e^{-2k\pi n}}{n}\)

\(\displaystyle =\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{e^{-2\pi kn}}{n}\)

\(\displaystyle =-\sum_{k=1}^{\infty}log(1-e^{-2\pi k})\)

Now, use the Theta identity mentioned above and take the log of both sides.

\(\displaystyle -log\left(\frac{{\pi}^{1/4}e^{\frac{\pi}{12}}}{\sqrt{2}\Gamma(3/4)}\right)\)

which is numerically the same as the posted solution.

Post Fri Jun 21, 2013 8:19 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Well Done, galactus!! We are moving closer to our target of 30 series. :D

Post Fri Jun 21, 2013 11:06 am
galactus User avatar
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Posts: 902
I reckon I should post the next one then.

Problem 23a:

This is an infinite product instead of summation.

Show that:

\(\displaystyle \displaystyle \prod_{k=2}^{\infty}\left(1-\frac{1}{k^{4}}\right)=\frac{\sinh(\pi)}{4\pi}\)

There is also one with a cube instead:

Problem 23b

\(\displaystyle \prod_{k=2}^{\infty}\left(1-\frac{1}{k^{3}}\right)=\frac{\cosh(\frac{\sqrt{3}\pi}{2})}{3\pi}\)

Perhaps there is a general form for these types of infinite products?. Maybe I should not have posted two for one problem, but I wanted to mention that perhaps there is a pattern with these types of products.

Let's say we have \(\displaystyle \prod_{k=2}^{\infty}\left(1-\frac{1}{k^{n}}\right)\)

If n is odd, then there is a closed form. I think it is:

\(\displaystyle \frac{1}{\prod_{k=1}^{n-1}\Gamma\left(2+(-1)^{k+1}e^{\frac{\pi ik}{n}}\right)}\)

Post Sat Jun 22, 2013 6:24 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

We may rewrite the product as

\[\prod_{k=2}^\infty \left( 1-\frac{1}{k^2}\right)\left( 1+\frac{1}{k^2}\right)\]

The first product is simply

\[
\begin{align*}
\prod_{k=2}^\infty \left( 1-\frac{1}{k^2}\right) &=\lim_{N\to \infty} \exp \sum_{k=2}^N \log \left( \frac{k^2-1}{k^2}\right) \\
&= \lim_{N\to \infty} \exp \left(-\log(2)+\log \left( 1+\frac{1}{N} \right)\right) \\
&= \frac{1}{2}
\end{align*}\]

The second product can be evaluated using Weierstrass's Product

\[
\begin{align*}
\prod_{k=2}^\infty \left( 1+\frac{1}{k^2}\right) &= \frac{1}{2}\prod_{k=1}^\infty \left(1+ \frac{i}{k}\right)\left( 1-\frac{i}{k}\right) \\
&= \frac{1}{2} \left( \frac{1}{i \Gamma(i)\Gamma(1-i)}\right) \\
&= \frac{\sinh(\pi)}{2\pi}
\end{align*}\]

Upon combining, we obtain
\[\prod_{k=2}^\infty \left( 1-\frac{1}{k^4}\right)=\frac{\sinh(\pi)}{4\pi}\]

Post Fri Jun 28, 2013 10:36 am
galactus User avatar
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Posts: 902
problem 24


\(\displaystyle \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{nk^{2}+2kn+n^{2}k}=7/4\)

Post Tue Aug 06, 2013 6:14 am
Random Variable Integration Guru
Integration Guru

Posts: 381
There is a formula (which I proved on another forum) that states for \(\displaystyle n >1\), \(\displaystyle \prod_{k=1}^{\infty} \Big(1- \frac{z^{n}}{k^{n}} \Big) = \prod_{k=0}^{n-1} \frac{1}{\Gamma[1-\exp(2 \pi i k/n)z]}\).

\(\displaystyle \prod_{k=2}^{\infty} \Big( 1- \frac{1}{k^{3}} \Big) = \lim_{z \to 1} \frac{1}{1-z^{3}} \prod_{k=1}^{\infty} \Big( 1- \frac{z^{3}}{k^{3}} \Big) =\lim_{z \to 1} \frac{1}{1-z^{3}} \frac{1}{\Gamma(1-z)} \frac{1}{\Gamma(1-e^{2 \pi i /3}z)} \frac{1}{\Gamma(1-e^{4 \pi i /3}z)}\)

\(\displaystyle = \frac{1}{\Gamma(1-e^{2 \pi i /3})} \frac{1}{\Gamma(1-e^{4 \pi i /3})} \lim_{z \to 1} \frac{1}{1-z^{3}} \frac{1}{\Gamma(1-z)}\)


\(\displaystyle \Gamma(1-e^{2 \pi i /3}) = \Gamma \Big( \frac{3}{2} - i \frac{\sqrt{3}}{2} \Big) = \Big( \frac{1}{2} - i \frac{\sqrt{3}}{2} \Big) \Gamma \Bigg( 1 - \Big( \frac{1}{2} + i \frac{\sqrt{3}}{2} \Big) \Bigg)\)

\(\displaystyle \Gamma(1-e^{4 \pi i /3}) = \Gamma \Big( \frac{3}{2} + i \frac{\sqrt{3}}{2} \Big) = \Big( \frac{1}{2}+ i \frac{\sqrt{3}}{2} \Big) \Gamma \Big( \frac{1}{2} + i \frac{\sqrt{3}}{2} \Big)\)


So \(\displaystyle \prod_{k=2}^{\infty} \Big( 1- \frac{1}{k^{3}} \Big) = \frac{1}{\pi} \cosh \left(\frac{\sqrt{3} \pi}{2} \right) \lim_{z \to 1} \frac{1}{1-z^{3}} \frac{1}{\Gamma(1-z)} = \frac{1}{\pi} \cosh \left(\frac{\sqrt{3} \pi}{2} \right) \lim_{z \to 0} \frac{1}{1-(z+1)^{3}} \frac{1}{\Gamma(-z)}\)

\(\displaystyle = \frac{1}{\pi} \cosh \left(\frac{\sqrt{3} \pi}{2} \right) \lim_{z \to 0} \frac{1}{-3z-3z^{2}-z^{3}} \frac{1}{\frac{-1}{z} + \mathcal{O}(1)} = - \frac{1}{ 3 \pi} \cosh \left(\frac{\sqrt{3} \pi}{2} \right) \lim_{z \to 0} \frac{1}{z} \frac{1}{1+z+\mathcal{O}(z^{2})} \Big( -z + \mathcal{O(z^{2})} \Big)\)

\(\displaystyle = - \frac{1}{3 \pi} \cosh \left(\frac{\sqrt{3} \pi}{2} \right) \lim_{z \to 0} \Big(1-z+\mathcal{O}(z^{2}) \Big) \Big(-z+\mathcal{O}(z^{2}) \Big) = \frac{1}{3 \pi} \cosh \left(\frac{\sqrt{3} \pi}{2} \right) \lim_{z \to 0} \Big( 1 + \mathcal{O} (z^{2}) \Big) = \frac{\cosh \left(\frac{\sqrt{3} \pi}{2}\right)}{3 \pi}\)


http://www.mathhelpboards.com/f28/infin ... #post26811

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