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I've been trying to prove, by arithmetical means, that

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

without success.

When I say "by arithmetical means" I mean to say, go from the left to the right expression just by symbolic manipulation.

Can anyone devise a way of doing this?

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you can find in [wolfram alpha][1] more series that are equal to this one, but it's not necessarily true that there is an algebraic way to show that. [1]: wolframalpha.com/input/… –  Dennis Gulko Jan 17 '12 at 11:38
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I seriously suspect there's an "arithmetical way" to do it. It is fairly obvious that one cannot transform one term to the other (or even a constant number of terms from the right to one term on the left), since the LHS converges much faster. –  aelguindy Jan 17 '12 at 11:45
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@Neves: that's exactly my point: you almost certainly can't. –  Dennis Gulko Jan 17 '12 at 11:48
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If you write it like $\sum (k^2 \binom{2k}{k} )^{-1}$ you find something here (eq 10)! –  draks ... Jan 17 '12 at 13:34
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@Emmand, see Sasha answer. –  Neves Jan 17 '12 at 14:44
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4 Answers 4

up vote 17 down vote accepted

Andreas gave you a good hint : the 'central binomial coefficient'. The generic term for this (widely studied) family of series is 'Central binomial sums' (series).

Your formula is a special case of the more general :
$$2(\arcsin(x))^2=\sum_{m=1}^{\infty} \frac{(2x)^{2m}}{m^2\binom{2m}{m}}$$

with $x=1/2$ and since $\arcsin(1/2)=\pi/6$ you'll get your answer.

You may find many formulas of this kind for example in Sprugnoli's 'Sums of reciprocals of the central binomial coefficients'.

Borwein and Broadhurst studied much this kind of series (very interesting reading by the way!) :

At the end of 'Pi and the AGM' the Borweins propose to prove the general formula using : $$x \frac{d}{dx}(\arcsin\ x)^2=\frac{2x \arcsin\ x}{\sqrt{1-x^2}}$$

and the fact that both $\displaystyle f(x)= \frac{\arcsin\ x}{\sqrt{1-x^2}}$ and $\displaystyle F(x)=\frac{1}{2x}\sum_{m=1}^{\infty} \frac{(2x)^{2m}}{m\binom{2m}{m}}$
satisfy the differential equation : $(1-x^2)f'=1+xf$

Perhaps not the direct proof you wished... Euler was probably the first to discover this formula as well as the other expressions of $\zeta(2n)$ (Euler's contributions to $\pi$ formulae).

Let's add that the formula proposed by the OP was part of Apery's famous proof of the irrationality of $\zeta(3)$ since he also proved the irrationality of $\zeta(2)$ using this formula (van der Poorten (1979) 'A proof that Euler missed..'. Compare the formula for $\zeta(3)$ there!).

Very nice stuff indeed!!

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I would have expected that WolframAlpha would give a hint on that. But still: Very nice. –  draks ... Jan 17 '12 at 23:01
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Adding to the last paragraph of Raymond Manzoni's answer. The given equality can be derived from the finite version $$ 2\sum_{n=1}^{N}\frac{(-1)^{n-1}}{n^{2}}+\sum_{k=1}^{N}\frac{(-1)^{N+k-1}}{ k^{2}\dbinom{N}{k}\dbinom{N+k}{k}}=3\sum_{n=1}^{N}\frac{1}{n^{2}\dbinom{2n}{n}}.\tag{1} $$

In the footnote 4 of Alf van der Poorten's article A proof that Euler missed ... Apéry's proof of the irrationality of $\zeta (3)$, the author states that the following identity $$ \zeta (2):=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6} =3\sum_{n=1}^{\infty }\frac{1}{n^{2}\dbinom{2n}{n}}\tag{2} $$ may be proved by slightly varying the argument in section 3 - multiply by $ (-1)^{n-1}$ instead of dividing by $n$. In this section 3 the equivalent one for $\zeta (3)$ is proved $$ \zeta (3):=\sum_{n=1}^{\infty }\frac{1}{n^{3}}=\frac{5}{2}\sum_{n=1}^{\infty } \frac{(-1)^{n-1}}{n^{3}\dbinom{2n}{n}},\tag{3} $$ as a consequence of$^1$ $$ \sum_{n=1}^{N}\frac{1}{n^{3}}=\frac{5}{2}\sum_{k=1}^{N}\frac{\left( -1\right) ^{k-1}}{k^{3}\dbinom{2k}{k}}+\sum_{k=1}^{N}\frac{\left( -1\right) ^{k-1}}{2k^{3}\dbinom{N}{k}\dbinom{N+k}{k}},\tag{4} $$ letting $N\rightarrow \infty $. I adapted the computation as indicated and obtained $(1)$. Since the second term on the left vanishes, as $N\rightarrow \infty $, we get the given equality in the form

$$ \sum_{n=1}^{\infty }\frac{1}{n^{2}}=2\sum_{n=1}^{\infty }\frac{ (-1)^{n-1}}{n^{2}}=3\sum_{n=1}^{\infty }\frac{1}{n^{2}\dbinom{2n}{n}}.\tag{5} $$


$^{1}$One of the intermediate sums can be written as $$ \sum_{k=1}^{n-1}(-1)^{k}n\left( \varepsilon _{n,k}-\varepsilon _{n-1,k}\right) =\frac{1}{n^{2}}-\frac{2(-1)^{n-1}}{n^{2}\dbinom{2n}{n}},\tag{6} $$

where $$ \varepsilon _{n,k}=\frac{1}{2}\frac{\left( k!\right) ^{2}(n-k)!}{k^{3}(n+k)!} =\frac{1}{2k^{3}\dbinom{n+k}{k}\dbinom{n}{k}}.\tag{7} $$

Instead of dividing $(6)$ by $n$ as a step to obtain $(4)$, if we multiply by $(-1)^{n-1}$ we get $$ \sum_{k=1}^{n-1}(-1)^{k+n-1}n\left( \varepsilon _{n,k}-\varepsilon _{n-1,k}\right) =\frac{(-1)^{n-1}}{n^{2}}-\frac{2}{n^{2}\dbinom{2n}{n}}.\tag{8} $$ After further manipulations I got $(1)$.

Note: the LHS of $(1)$ is the diagonal sequence $c_{N,N}^{\prime }$ of the double sequence $c_{n,k}^{\prime }$ defined by the formula $5^{\prime }$ in section 6 of the mentioned article.

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First, let's compute a slightly simpler sum $$ \begin{align} \sum_{k=1}^\infty\frac{\Gamma(k)^2}{\Gamma(2k)}(2x)^{2k-1} &=\sum_{k=1}^\infty\mathrm{B}(k,k)(2x)^{2k-1}\\ &=\sum_{k=1}^\infty\int_0^1t^{k-1}(1-t)^{k-1}(2x)^{2k-1}\mathrm{d}t\\ &=\int_0^1\frac{2x}{1-4x^2t(1-t)}\mathrm{d}t\\ &=\frac{1}{2x}\int_0^1\frac{1}{t^2-t+\frac{1}{4x^2}}\mathrm{d}t\\ &=\frac{1}{2x}\int_0^1\frac{1}{(t-\alpha)(t-1+\alpha)}\mathrm{d}t\text{ where }2\alpha-1=\sqrt{1-\frac{1}{x^2}}\\ &=\frac{1}{2x}\frac{1}{2\alpha-1}\int_0^1\left(\frac{1}{t-\alpha}-\frac{1}{t-1+\alpha}\right)\mathrm{d}t\\ &=\frac{1}{2\sqrt{x^2-1}}\left[\log\left(\frac{\alpha-1}{\alpha}\right)-\log\left(\frac{\alpha}{\alpha-1}\right)\right]\\ &=\frac{1}{\sqrt{x^2-1}}\log\left(\frac{\alpha-1}{\alpha}\right)\text{ where }\frac{\alpha-1}{\alpha}=\left(\sqrt{1-x^2}+ix\right)^2\\ &=\frac{-2i}{\sqrt{1-x^2}}i\,\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\\ &=\frac{2}{\sqrt{1-x^2}}\sin^{-1}(x)\tag{1} \end{align} $$ Integrating both sides of $(1)$ yields $$ \frac12\sum_{k=1}^\infty\frac{\Gamma(k)^2(2x)^{2k}}{\Gamma(2k+1)} =\left[\sin^{-1}(x)\right]^2\tag{2} $$ Plugging $x=\frac12$ into $(2)$ gives $$ \sum_{k=1}^\infty\frac{(k-1)!^2}{(2k)!}=\frac{\pi^2}{18}\tag{3} $$ In this answer, it is shown that $$ \sum_{k=1}^\infty\frac{1}{k^2}=\zeta(2)=\frac{\pi^2}{6}\tag{4} $$ Combining $(3)$ and $(4)$ yields $$ \sum_{k=1}^\infty\frac{(k-1)!^2}{(2k)!}=\frac13\sum_{k=1}^\infty\frac{1}{k^2}\tag{5} $$

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Awesome, Man! I really like this. –  user9413 May 31 '12 at 16:40
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In your previous question you asked to evaluate the left-hand-side, which gave $\frac{\pi^2}{18}$.

The left-hand-side is $\frac{1}{3} \sum_{k=1}^\infty \frac{1}{k^2} = \frac{1}{3} \zeta(2) = \frac{1}{3} \frac{\pi^2}{6} = \frac{\pi^2}{18}$.

Here $\zeta(2)$ stands for the Riemann zeta-function.

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it wasn't my question, but inspired this one. I just wonder if there is a sequence of arithmetic steps leading from LHS to RHS. –  Neves Jan 17 '12 at 14:32
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Proving rhs equivalent to lhs algebraically can be done by finding a Wilf-Zeilberger pair. The book "A=B" provides details. –  Sasha Jan 17 '12 at 14:51
    
@Sasha, thank you so much for sharing such resources. The concept is quite interesting indeed. –  Emmad Kareem Jan 17 '12 at 16:31
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