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?

  • 3
    $\begingroup$ What source are you using as a reference? I do not recall actually having to examine the $e_{n,k}$ directly, but rather some related sums involving the individual terms in $e_{n,k}$. In any case, a good readable account is in "A PROOF THAT EULER MISSED. AP´ERY’S PROOF OF THE IRRATIONALITY OF $\zeta(3)$, AN INFORMAL REPORT", by van der Poorten, available at his website, www.maths.mq.edu.au/~alf/45.pdf $\endgroup$ Mar 6, 2011 at 18:03
  • $\begingroup$ A link to the article by Alf van der Poorten is this ift.uni.wroc.pl/~mwolf/Poorten_MI_195_0.pdf . The quantity you mention appears in section 3. $\endgroup$ Mar 6, 2011 at 18:20
  • $\begingroup$ @Americo: Not really, the sum itself is not needed. In section 3, the individual terms that make up $e_{n,k}$ are used in other sums. (Also, note that the version I linked to is slightly more updated that the printed article.) $\endgroup$ Mar 6, 2011 at 18:45
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    $\begingroup$ @Andres: Concerning the sum that's right it is not needed for proving the irrationality of $\zeta (3)$. Anyhow the sum I referred to is $\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}\rightarrow 0 $ as $N\rightarrow \infty $. According to my calculations $\left\vert \sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}% \right\vert <\frac{1}{2(N+1)}$. $\endgroup$ Mar 6, 2011 at 19:37
  • $\begingroup$ @Americo: You are right (and your estimate is easier than what I was getting). $\endgroup$ Mar 6, 2011 at 19:41

3 Answers 3


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 $k$.

  • $\begingroup$ @Didier: In the edit, which is the technique/argument one should use to find $m_{\text{min}}\approx n/\sqrt{2}$ ? $\endgroup$ May 17, 2011 at 17:32
  • $\begingroup$ @Américo: One can estimate the ratio $b_{n,m}/b_{n,m+1}$ and compare it to $1$. $\endgroup$
    – Did
    May 17, 2011 at 17:33
  • $\begingroup$ @Didier: I verified that for $m\approx n/\sqrt{2}$ the sequence $b_{n,m}/b_{n,m+1}\rightarrow 1$ (as $n\rightarrow \infty $). $\endgroup$ May 17, 2011 at 18:05
  • $\begingroup$ @Américo: Sorry I was not more explicit but you fix $n$ and you show that $b_{n,m}/b_{n,m+1}$ is larger than $1$ for $m\le m(n)$ and smaller than $1$ for $m>m(n)$, where $m(n)\sim n/\sqrt2$. Hence the sequence $(b_{n,m})_m$ (for a fixed $n$) is nonincreasing then nondecreasing, which is all you need to conclude. $\endgroup$
    – Did
    May 17, 2011 at 22:17
  • $\begingroup$ @Didier: Thanks! Now makes sense. $\endgroup$ May 17, 2011 at 22:22

Update: edited text and formatting.

We can write the sum $e_{n,k}$ in the form (see section 4 of this article by Alf van der Poorten)

$$e_{n,k}=\sum_{m=1}^{k}\frac{(-1)^{m-1}}{2m^{3}\dbinom{n}{m}\dbinom{n+m}{m}}, \quad (1\leq k\leq n).$$

Let $u_{n,m}=m^{3}\binom{n}{m}\binom{n+m}{m}$. If $1=m\leq n$, then $u_{n,m}=n\left( n+1\right)$. To show that $u_{n,m}>n(n+1)$ for $1<m\leq n$ we consider the following two cases:

  • if $1<m=n$, then $u_{n,m}=n^{3}\binom{2n}{n}>n(n+1)$;

  • if $1<m\leq n-1$, then $m^{3}\binom{n}{m}\geq m^{3}\binom{n}{1}=m^{3}n>n$ and $\binom{n+m}{m}\geq \binom{n+m}{1}=n+m>n+1$. Hence $u_{n,m}>n(n+1)$.

Hence, for $1<k\leq n$, we get:

$$\begin{eqnarray*} \left\vert e_{n,k}\right\vert &=&\left\vert \sum_{m=1}^{k}\frac{(-1)^{m-1}}{% 2m^{3}\binom{n}{m}\binom{n+m}{m}}\right\vert \leq \sum_{m=1}^{k}\left\vert \frac{(-1)^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\right\vert\leq \sum_{m=1}^{n}\frac{1}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\\&<&\sum_{m=1}^{n}\frac{1}{2n(n+1)}=\frac{n}{2n(n+1)}<\frac{1}{n}. \end{eqnarray*}$$

For $k=1$, we get $\left\vert e_{n,1}\right\vert =\frac{1}{2n\left( n+1\right) }% \leq \frac{1}{2(n+1)}<\frac{1}{n}$. Thus for every integer $1\leq k\leq n$, we proved that $\left\vert e_{n,k}\right\vert <\frac{1}{n}$, which implies that $e_{n,k}$ converges uniformly in $k$ to $0$.


If $k$ is fixed, all the terms are $O(1/n^2)$, and more over this is an alternating sum, so the entire sum is $O(1/n^2)$. For the uniformity I guess a more delicate argument is needed, but probably Stirling's approximation is your friend here.


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