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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?

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@Chandrasekhar: No, it means the double factorial. – Zev Chonoles Oct 31 '11 at 20:01
The above series for $\frac{\pi}{4}$ should be called the Madhava formula, or the Leibniz formula, or the Gregory formula, or combinations of some or all the names (Madhava came first). No Machin. – André Nicolas Oct 31 '11 at 20:37
@Joren Why don't you click the link in Zev's comment and find out? ;) – Srivatsan Nov 1 '11 at 15:11
Would the downvoter care to comment? – robjohn Dec 13 '12 at 12:01
A 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. – robjohn Aug 4 '14 at 6:58
up vote 48 down vote accepted

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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That'll do it! Thanks for the reference. – robjohn Oct 31 '11 at 19:39
See also this related question. – Mike Spivey Oct 31 '11 at 19:40
I was able to use the formula for the integral of an odd power of sine to come up with a fourth method to sum my series. Unfortunately, it is too long for the margin. – robjohn Nov 1 '11 at 15:07
@robjohn: Your comments on my answers keep being too long for the margin... ;) – Mike Spivey Nov 1 '11 at 15:25

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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+1: I figured there was a way to adapt T..'s argument in the question I linked to. – Mike Spivey Oct 31 '11 at 21:06
@MikeSpivey: Thanks! I quickly looked over the paper you linked to after posting this. It is worth noting that the switching of the order is pretty much the key thing in every proof there. – Eric Naslund Oct 31 '11 at 21:11
The bounds on the second last integral should be from $u=1$ to $\infty$. – SL2 Oct 31 '11 at 21:12
@SL2 Corrected! thanks – Eric Naslund Oct 31 '11 at 21:13
(You might want to add "dx" to your integrals, BTW.) – Mike Spivey Oct 31 '11 at 21:16

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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Nice. I like seeing more that one way to skin a cat. This is the same function as given in Mike Spivey's post since $\tan(x)=\frac{\sin(x)}{\sqrt{1-\sin^2(x)}}$. – robjohn Oct 31 '11 at 19:42

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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Why was the result suddenly halved? – J. M. Nov 1 '11 at 15:04
@JM: thanks for catching the typo. – robjohn Nov 1 '11 at 15:13
@robjohn: It is worth noting that this is really just using the Beta Function in disguise. First, $$\int_0^\frac{\pi}{2} \sin^{2k-1}(x)dx=\frac{1}{2}\int_0^{\pi} \sin^{2k-1}(x)dx=\int_0^\frac{\pi}{2} \sin^{2k-1}(2x)dx=2\int_0^\frac{\pi}{2} \sin^{2k-1}(x)\cos^{2k-1}(x)dx.$$ Letting $u=\sin^2(x)$, we have $du=2\sin(x)\cos(x)$, so that this is $$\int_0^1 x^k (1-x)^k dx.$$ In other words, this solution is identical to the one above, but a change of variables has taken place first. In general we can write the beta function as $$\text{B}(x,y)=2\int_0^\frac{\pi}{2}\sin^{2x-1}(x)\cos^{2y-1}(x)dx.$$ – Eric Naslund Nov 1 '11 at 17:16
@Eric: I was actually about to give a proof (in a comment to Mike's answer) very similar to yours, using the Beta function, but you posted first, so I switched to using the other identity. When I saw that it came to $\int_0^1\frac{2\;\mathrm{d}t}{1+t^2}$, I figured it was still similar to your answer. – robjohn Nov 1 '11 at 17:31
@Eric: you probably meant $$\textrm{B}(x,y)=2\int_0^\frac{\pi}{2}\sin^{2x-1}(t)\;\cos^{2y-1}(t)\;\textrm{d‌​}t$$ – robjohn Nov 1 '11 at 17:35

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