The sum of the series $ \frac{\pi}{2}=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}\tag{1} $ can be derived by accelerating the Gregory Series $ \frac{\pi}{4}=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2} $ using Euler's Series Transformation. Mathematica is able to sum $(1)$, so I assume there must be some method to sum the series in $(1)$ directly; what might that method be?
How to sum this series for $\pi/2$ directly?
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10A third downvote without comment. I derived the series using the Euler Series Transform and even posted an answer. I believe I have shown what I have done and my subsequent effort, if that is what is bothering people. Of course, perhaps something else is wrong with my question, but since no one is commenting on the downvotes, I can't really do anything about them. – 2014-08-04
5 Answers
First, $(2k+1)!! = (2k+1)(2k-1) \cdots (1) = \frac{(2k+1)!}{(2k)(2(k-1)) \cdots 2(1)} = \frac{(2k+1)!}{2^k k!}.$
So your sum can be rewritten as
$\sum_{k=0}^\infty\frac{k! \, k! \, 2^k }{(2k+1)!} = \sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}.$
Variations of the sum of reciprocals of the central binomial coefficients have been well-studied. For example, this paper by Sprugnoli (see Theorem 2.4) gives the ordinary generating function of $a_k = \frac{4^k}{(2k+1)}\binom{2k}{k}^{-1}$ to be $A(t) = \frac{1}{t} \sqrt{\frac{t}{1-t}} \arctan \sqrt{\frac{t}{1-t}}.$
Subbing in $t = 1/2$ says that $\sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}} = 2 \arctan(1) = \frac{\pi}{2}.$
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5@robjohn: Your comments on my answers keep being too long for the margin... ;) – 2011-11-01
We can prove this identity, as well as the corresponding power series identities by using a relation with the Beta function. Rearranging as done in Mike Spivey's answer we are looking at $ \sum_{k=0}^\infty\frac{k! k! 2^k}{(2k+1)!}$ Using induction or a Beta Function identity, we can show that $\int_0^1 x^{k}(1-x)^k=\frac{k!k!}{(2k+1)!}.$ Hence your sum becomes
$ \sum_{k=0}^\infty 2^k \int_0^1 x^{k}(1-x)^k=\int_0^1 \left(\sum_{k=0}^\infty 2^k x^k (1-x)^k\right)dx.$
Notice that since $0\leq x\leq 1$, $x(1-x)\leq \frac{1}{4}$ and the series converges absolutely. Summing gives
$=\int_0^1 \frac{1}{1-2x(1-x)}dx=\int_0^1 \frac{1}{x^2+(1-x)^2}dx$ Substituting $u=\frac{1}{x}$, and then $v=u-1$, we see that this integral is equal to $\int_1^\infty \frac{1}{1+(u-1)^2}du=\int_0^\infty \frac{1}{1+v^2}dv=\frac{\pi}{2},$ as desired.
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0I saw the Beta integral on the page that Mike Spivey linked to, and I was working on just this method. Thanks for a third method! – 2011-10-31
I had intended for this to be a comment to Mike Spivey's answer, but it is too long.
One of the answers to the related question mentions a result equivalent to $ \int_0^\frac{\pi}{2}\sin^{2k+1}(x)\;\mathrm{d}x=\frac{2k}{2k+1}\frac{2k-2}{2k-1}\cdots\frac{2}{3}=\frac{1}{2k+1}\frac{4^k}{\binom{2k}{k}}\tag{1} $ Using $(1)$, my sum becomes $ \begin{align} \sum_{k=0}^\infty\frac{k!}{(2k+1)!!} &=\sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}\\ &=\sum_{k=0}^\infty\int_0^\frac{\pi}{2}\sqrt{2}\left(\frac{\sin(x)}{\sqrt{2}}\right)^{2k+1}\mathrm{d}x\\ &=\sqrt{2}\int_0^\frac{\pi}{2}\frac{\left(\frac{\sin(x)}{\sqrt{2}}\right)}{1-\left(\frac{\sin(x)}{\sqrt{2}}\right)^2}\;\mathrm{d}x\\ &=\int_0^\frac{\pi}{2}\frac{2\,\sin(x)}{2-\sin^2(x)}\;\mathrm{d}x\\ &=\int_\frac{\pi}{2}^0\frac{2\;\mathrm{d}\cos(x)}{1+\cos^2(x)}\\ &=\frac{\pi}{2} \end{align} $
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1@robjohn: Oops!! I chose the wrong variable of integration there! Ya, I just wanted to point out here that every solution so far, in principle, is the same, which is kinda interesting. (I can't really tell with Sasha's, since a proof of the main identity is not cited) – 2011-11-01
Notice that for $c_k = \frac{k!}{(2k+1)!!}$ the ratio of successive terms $\frac{c_{k+1}}{c_k} = \frac{k+1}{2k +3} = \frac{1}{2} \frac{k+1}{k+3/2}$.
This means that the series is hypergeometric with the value ${}_2 F_1(1, 1, \frac{3}{2}, \frac{1}{2})$.
This particular Gaussian hypergeometric is elementary: $ {}_2 F_1(1, 1, \frac{3}{2}, x) = \frac{\arcsin\left(\sqrt{x}\right)}{\sqrt{1-x} \sqrt{x}} $ Upon substitution of $x=\frac{1}{2}$ we recover the result $ 2 \arcsin(\frac{1}{\sqrt{2}}) = \frac{\pi}{2}$.
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9Nice. I like seeing more that one way to skin a cat. This is the same function as given in Mike Spivey's post since $\tan(x)=\frac{\sin(x)}{\sqrt{1-\sin^2(x)}}$. – 2011-10-31