TITLE: Proving a sequence involved in Apéry's proof of the irrationality of $\zeta(3)$, converges
QUESTION [28 upvotes]: I am trying to understand Apery's 1978 proof that $\zeta(3) = \displaystyle \sum_{n=1}^\infty \frac{1}{n^3}$ is irrational. The idea behind the proof is to find an 'accelerated' series for $\zeta(3)$ which converges too fast to $\zeta(3)$, thus proving that $\zeta(3)$ cannot be rational. In particular, a particular quantity is defined:
$$e_{n,k} = \displaystyle \sum_{m=1}^k \frac{(-1)^{m-1} (m!)^2 (n-m)!}{2m^3 (n+m)!},\quad k \leq n.$$
The key is to show that $\displaystyle \lim_{n \rightarrow \infty} e_{n,k} = 0$ uniformly in $k$, and I have no idea why this sum converges to 0. Any ideas?

REPLY [22 votes]: By the triangular inequality, $|e_{n,k}|\leqslant\frac12 a_n$ with
$$
a_n=\sum_{m=1}^nb_{n,m},\qquad b_{n,m}=\frac{(m!)^2(n-m)!}{m^3(n+m)!},
$$
hence the desired uniform convergence holds as soon as $a_n\to0$. 
To prove that $a_n\to0$, note that
$$
b_{n,m}=\frac{(m-1)!(m-1)!(n-m)!}{m(n+m)!}\leqslant\frac{(m-1)!(m-1)!(n-m)!}{(n+m)!},
$$
and use twice the fact that $i!j!\leqslant (i+j)!$ for every nonnegative integers $i$ and $j$. This yields
$$
b_{n,m}\leqslant\frac{(n+m-2)!}{(n+m)!}=\frac1{(n+m-1)(n+m)}=\frac1{n+m-1}-\frac1{n+m}.
$$
Thus $a_n$ is bounded by a telescoping sum, namely
$$
a_n\leqslant\sum_{m=1}^n\left(\frac1{n+m-1}-\frac1{n+m}\right)=\frac1{n}-\frac1{2n}=\frac1{2n},
$$
This shows that $|e_{n,k}|\leqslant\frac1{4n}$ for every $n\geqslant1$ and uniformly over $k\leqslant n$, hence the proof is complete.
Edit One can refine the estimates above, taking into account the variations of the sequence  $(b_{n,m})_{1\leqslant m\leqslant n}$ for some fixed $n$, which is nonincreasing on $1\leqslant m\leqslant m_n$ for some $m_n\approx n/\sqrt2$ and nondecreasing on $m_n\leqslant m\leqslant n$. Thus, $|e_{n,k}|\leqslant\frac12  b_{n,1}+\frac12 b_{n,n}$. Hence, $n^2e_{n,k}\to\frac12 $ uniformly over $k$.