# Ramanujan Series

I have the following question involving the series of $1/\pi^3$: Can we find such expansions by using the one for $1/\pi^3$ with $1/\pi^4$ or $1/\pi^n$ etc?

Note that $$\frac{1}{32}\sum_{n=0}^\infty \frac{\left(\frac{1}{2}\right)^7_n(168n^3+76n^2+14n+1)}{2^{10n} n!^5}=\frac{1}{\pi^3}.$$

Can we extend the above formula for $1/\pi^n$?

Also, I had the following question: we know that $n!=1\times2\times\cdots\times(n-1)\times n$. Can we find $\log(n!)$? I think, $\log(n!) = \log(1\cdot2\cdot3\cdots n)=\log1+\log2+\log3+\cdots+\log n$. Am I right or wrong? - I don't know. Please explain.

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$(1/32)\sum_{n=0}^{\infty}{(1/2)^7\over(n!)^7}2^{6n}[168n^3+76n^2+14n+1]=(1/\pi)‌​^3$ is what I think you "observe" - did I get that right? Did you observe this yourself? Is it from Ramanujan? What do you mean by "extend" the formula? – Gerry Myerson Oct 21 '11 at 10:48
@Gerry: I just tried that numerically in Mathematica, and it doesn't seem to be true. N[Sum[(1/2)^7/(n!)^7 * 2^(6n) ( 168n^3 + 76n^2 + 14n + 1),{n,0,Infinity}] / 32, 20] gives 18.362683970772570548. – Hans Lundmark Oct 21 '11 at 10:50
@Hans, thanks. Maybe the $2^{6n}$ is supposed to be in the denominator. Maybe OP will return and clarify. – Gerry Myerson Oct 21 '11 at 10:57
@Gerry: I fixed the equation - three major parts of it were wrong: the $\left(\frac{1}{2}\right)^7_n$ I do not think is supposed to be interpreted as a power; the $2^{6n}$ should be $32^{2n}$, or $2^{10n}$; and the power of $n!$ should be $5$, not $7$. See equation 125 at MathWorld's page on Pi Formulas. – anon Oct 21 '11 at 10:59
@anon, thanks. I take it $(1/2)^7_n$ is the 7th power of a Pochhammer (falling factorial) symbol? Well, I guess I should check your link. – Gerry Myerson Oct 21 '11 at 11:04