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How could one prove that

$$\sum_{k=0}^\infty \frac{2^{1-k} (3-25 k)(2 k)!\,k!}{(3 k)!} = -\pi$$

I’ve seen similar series, but none like this one…

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It seems irreducible in current form, and I have no idea as to what kind of transformation might aid in finding proof of this.

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- Could this approximation be made simpler ? Solve $n!=a^n 10^k$

Use Beta function, I guess… for $k \ge 1$,

$$

\int_0^1 t^{2k}(1-t)^{k-1}dt = B(k,2k+1) = \frac{(k-1)!(2k)!}{(3k)!}

$$

So write

$$

f(x) = \sum_{k=0}^\infty \frac{2(3-25k)k!(2k)!}{(3k)!}x^k

$$

and compute $f(1/2)$ like this:

$$\begin{align}

f(x) &= 6+\sum_{k=1}^\infty \frac{(6-50k)k(k-1)!(2k)!}{(3k)!} x^k

\\ &= 6+\sum_{k=1}^\infty (6-50k)k x^k\int_0^1 t^{2k}(1-t)^{k-1}dt

\\ &= 6+\int_0^1\sum_{k=1}^\infty (6-50k)k x^k t^{2k}(1-t)^{k-1}\;dt

\\ &= 6+\int_0^1 \frac{4t^2x(14t^3x-14t^2x-11)}{(t^3x-t^2x+1)^3}\;dt

\\ f\left(\frac{1}{2}\right) &=

6+\int_0^1\frac{16t^2(7t^3-7t^2-11)}{(t^3-t^2+2)^3}\;dt

= -\pi

\end{align}$$

……

of course any calculus course teaches you how to integrate a rational function…

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