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?
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!!
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.
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 $$ \sum_{k=1}^\infty\frac{(k-1)!^2}{(2k)!}=\frac13\sum_{k=1}^\infty\frac{1}{k^2}\tag{5} $$
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.
