# Yet another $\sum = \pi$. Need to prove.

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…

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

#### Solutions Collecting From Web of "Yet another $\sum = \pi$. Need to prove."

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…