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Typo in formula
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R.P.
R.P.
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Ramanujan's famous pi formula states that \begin{equation} \frac{1}{\pi}=\frac{2\sqrt{2}}{99^2}\sum_{k=0}^{\infty}\frac{4k!}{k!^4}\frac{26390k+1103}{396^{4k}} \end{equation}\begin{equation} \frac{1}{\pi}=\frac{2\sqrt{2}}{99^2}\sum_{k=0}^{\infty}\frac{(4k)!}{k!^4}\frac{26390k+1103}{396^{4k}} \end{equation} How can one prove this? If the proof is too long for this site, you can reference any article containing the proof.

Ramanujan's famous pi formula states that \begin{equation} \frac{1}{\pi}=\frac{2\sqrt{2}}{99^2}\sum_{k=0}^{\infty}\frac{4k!}{k!^4}\frac{26390k+1103}{396^{4k}} \end{equation} How can one prove this? If the proof is too long for this site, you can reference any article containing the proof.

Ramanujan's famous pi formula states that \begin{equation} \frac{1}{\pi}=\frac{2\sqrt{2}}{99^2}\sum_{k=0}^{\infty}\frac{(4k)!}{k!^4}\frac{26390k+1103}{396^{4k}} \end{equation} How can one prove this? If the proof is too long for this site, you can reference any article containing the proof.

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David Roberts ♦
David Roberts ♦
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russian bot
russian bot
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Ramanujan's infinite sum for pi

Ramanujan's famous pi formula states that \begin{equation} \frac{1}{\pi}=\frac{2\sqrt{2}}{99^2}\sum_{k=0}^{\infty}\frac{4k!}{k!^4}\frac{26390k+1103}{396^{4k}} \end{equation} How can one prove this? If the proof is too long for this site, you can reference any article containing the proof.