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$$ \frac{2 \sqrt 2}{99^2} \sum_{n=1}^{\infty} \frac{(4n)!(1103+26390n)}{(4^n 99^n n!)^4} $$

How to calculate? I can't think how to. Wolfram said

$$ \frac{1}{\pi} - \frac{2206\sqrt2}{99^2} $$

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    Your expression is not very readable. – 2017-02-25
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    Do you mean $\frac{2√2}{99^2} \sum_{n=1}^{∞} \frac{(4n)!(1103+26390n)}{(4^n 99^n n!)^4)}$ ? – 2017-02-25
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    @frederick99: complete with mismatch of parentheses – 2017-02-25
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    Please use `\sqrt` instead of the non-ASCII character `√`. It allows for better portability. It also formats better: $\sqrt2$ vs $√2$. – 2017-02-25
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    what does Wolfram Alpha say? – 2017-02-25
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    This smells of Ramanujan... – 2017-02-25
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    I have the strong feeling *Ramanujan* has something to do with it. – 2017-02-25
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    @Chappers: as a matter of fact... https://crypto.stanford.edu/pbc/notes/pi/ramanujan.html – 2017-02-25
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    The derivation is heavily related with elliptic integrals. – 2017-02-25
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    see related http://math.stackexchange.com/a/434237/72031 and some theory in my blogs http://paramanands.blogspot.com/2012/03/modular-equations-and-approximations-to-pi-part-2.html – 2017-02-25

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This question is equivalent to showing that $$\frac1{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\!.$$

This formula for $\frac{1}{\pi}$ was discovered by Ramanujan, and for a discussion of its derivation, I suggest the following: Motivation for Ramanujan's mysterious $\pi$ formula.

Since your sum is readily seen to be the same as Ramanujan's formula only differing in the factorization of numbers and in that your sum begins at $k = 1$ instead of $k = 0$. Thus, $$\frac{2 \sqrt 2}{99^2} \sum_{n=1}^{\infty} \frac{(4n)!(1103+26390n)}{(4^n 99^n n!)^4}$$ is $\frac{1}{\pi}$ minus the $k=0$ term of Ramanujan's series, namely, $\frac{2 \cdot 1103 \sqrt 2}{99^2}.$