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Series Contest (SC3)

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Post Sun Dec 22, 2013 7:10 am
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

mathbalarka wrote:
Am I correct in thinking that this has a whole lot to do with thetas?


No, this does not involve elliptics. It is much simpler than that.

Post Wed Jan 29, 2014 9:33 am

Posts: 8
Shobhit wrote:
[BONUS]

\(\displaystyle \frac{4}{\pi}=\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n^3}{(n!)^3}\frac{6n+1}{4^n}\)

[NOTE] \(\displaystyle (a)_n = \frac{\Gamma(a+n)}{\Gamma(a)}\)

Here is a solution for the BONUS Problem using brute force:

Differentiating the original statement of Problem 8 with respect to \(\displaystyle k\) using the identity \(\displaystyle \frac{d}{dk}K(k)=\frac{E(k)-(1-k^2)K(k)}{k(1-k^2)}\) we get

\(\displaystyle \begin{align*}
4 \frac{1-2k^2}{\sqrt{1-k^2}}\sum_{j=0}^\infty \frac{\left(\frac{1}{2} \right)_j^3 j }{(j!)^3}(2kk')^{2j-1}&=\frac{8K(k)}{\pi^2}\left(\frac{E(k)-\left(1-k^2\right) K(k)}{k \left(1-k^2\right)} \right) \\
\implies \sum_{j=0}^\infty \frac{\left(\frac{1}{2} \right)_j^3 j }{(j!)^3}(2kk')^{2j} &= \frac{4K(k)}{\pi^2}\left\{\frac{E(k)-\left(1-k^2\right) K(k)}{(1-2k^2) } \right\}\tag{1}\\
\end{align*}\)

Multiply equation (1) by 6 and add it to the statement of problem 8:

\(\displaystyle \sum_{j=0}^\infty \frac{\left(\frac{1}{2} \right)_j^3 (6j+1) }{(j!)^3}(2kk')^{2j}=\frac{4K(k)^2}{\pi^2}+\frac{24K(k)}{\pi^2}\left\{\frac{E(k)-\left(1-k^2\right) K(k)}{(1-2k^2) } \right\}\tag{2}\)

Put \(\displaystyle k=k_3=\frac{\sqrt{3}-1}{2}\) and observe that \(\displaystyle 2k_3 k_3^{'}=\frac{1}{2}\) and \(\displaystyle K(k_3)=\frac{3^{1/4}\Gamma^3\left(\frac{1}{3} \right)}{2^{7/3}\pi}\). After some laborious simplifications we get

\(\displaystyle \begin{align*}
\sum_{j=0}^\infty \frac{\left(\frac{1}{2} \right)_j^3}{(j!)^3}\frac{6j+1}{4^j} &= \frac{4}{\pi} \tag{3}
\end{align*}\)

Post Tue May 03, 2016 7:53 am
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

Shobhit wrote:
Here is another problem for anybody interested:

Problem 9

\(\displaystyle \sum_{n=0}^\infty \frac{1}{F_{2n+1}+1}=\frac{1}{2}\sqrt5\)

where \(F_n\) is a Fibonacci Number.


Luckily, it can be evaluated without thetas and elliptics. Let $\phi=\frac{1+\sqrt{5}}{2}$ denote the golden ratio. Then consider the partial sum,

\begin{align*}
\sum_{n=0}^N\frac{1}{1+F_{2n+1}}&= \sum_{n=0}^N\frac{1}{1+\frac{\phi^{2n+1}+\phi^{-(2n+1)}}{\sqrt{5}}} \\
&= \sqrt{5} \sum_{n=0}^{N}\frac{\phi^{2n+1}}{\phi^{2(2n+1)}+\sqrt{5}\phi^{2n+1}+1} \\
&=\sqrt{5} \sum_{n=0}^{N}\frac{\phi^{2n+1}}{(\phi^{2n+1}+\phi)\left( \phi^{2n+1}+\frac{1}{\phi}\right)}\\
&= \sqrt{5} \sum_{n=0}^{N}\frac{\phi^{2n+1}}{(\phi^{2n}+1)\left( \phi^{2n+2}+1\right)} \\
&= \frac{\phi\sqrt{5}}{1-\phi^2}\sum_{n=0}^N\left(\frac{\phi^{2n}}{1+\phi^{2n}}-\frac{\phi^{2n+2}}{1+\phi^{2n+2}} \right) \\
&=\sqrt{5}\left(\frac{\phi^{2N+2}}{1+\phi^{2N+2}} -\frac{1}{2}\right)
\end{align*}

Let $N\to\infty$ to get $$\sum_{n=0}^\infty\frac{1}{1+F_{2n+1}}=\frac{\sqrt{5}}{2}$$

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