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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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    @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.2011-10-21

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