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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...

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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    $\begingroup$ This looks like a Generalized Hypergeometric Series evaluated at $n=\frac12$ (thus the $2^{-k}$ term in the numerator); it seems to bear a lot of similarity to some of Chudnovsky or Ramanujan's formulae for $\pi$. Have you tried writing the sum as a hypergeometric function and looking for any relevant identities? $\endgroup$ – Steven Stadnicki Jun 21 '13 at 18:48
  • $\begingroup$ @StevenStadnicki: It's almost certainly a GHS, but I can't prove it... $\endgroup$ – JohnWO Jun 21 '13 at 18:56
  • $\begingroup$ Related: math.stackexchange.com/questions/1652812/… $\endgroup$ – Jaume Oliver Lafont Feb 20 '16 at 22:39
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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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  • $\begingroup$ Why the rewrite of $f(x)$ such that $f(x) = 6+\sum\cdots$? $\endgroup$ – JohnWO Jun 21 '13 at 20:26
  • $\begingroup$ That Beta integral diverges when $k=0$. $\endgroup$ – GEdgar Jun 21 '13 at 20:30
  • $\begingroup$ Sorry, I'm unable to interpret your answer. $\endgroup$ – JohnWO Jun 21 '13 at 20:47
  • $\begingroup$ +1, this is about what I expected. (Also note that by the rational root theorem and a bit of trial-and-error, $t+1$ is a factor of the cubic in the denominator.) @JohnWO , what part of this is unclear to you? $\endgroup$ – Steven Stadnicki Jun 21 '13 at 21:34
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    $\begingroup$ @JohnWO: To detail a little GEdgar's fine answer : $\sum_{k=0}\cdots=6+\sum_{k=1}\cdots$ (to avoid later difficulties with $k=0$) $$f(x):=\sum_{k=1}^\infty \left(x\,t^2\,(1-t)\right)^k=\frac{t^2x(1-t)}{1-t^2\,x\,(1-t)}$$ $$xf(x)':=\sum_{k=1}^\infty k\left(x\,t^2\,(1-t)\right)^k=\frac{t^2x(1-t)}{(1-t^2\,x\,(1-t))^2}$$ $$x(xf(x)')':=\sum_{k=1}^\infty k^2\left(x\,t^2\,(1-t)\right)^k=\frac{t^2x(1-t)(1+t^2\,x\,(1-t))}{(1-t^2\,x\,(1-t))^3}$$ so that (rewriting $(1-t)^{k-1}=\frac{(1-t)^k}{1-t}$) : $\endgroup$ – Raymond Manzoni Jun 21 '13 at 21:35

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