12
$\begingroup$

Here is a symmetric rational double series giving Apery's constant:

$$ \frac13 \sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \displaystyle \frac{(i-1)! (j-1)!}{(i+j)!} H_{i+j} = \zeta(3) $$

where $\displaystyle H_{n}:=\sum_{1}^{n} \frac{1}{k}$, $n=1,2,\cdots,$ are the harmonic numbers.

How would you prove it?

Edit. In 2005, I sent this result to Wolfram MathWorld (see equation 25), I built it as an echo of

$$ \sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \displaystyle \frac{(i-1)! (j-1)!}{(i+j)!} = \zeta(2). $$

See Jack's pretty answer and see my answer below.

$\endgroup$
  • 1
    $\begingroup$ Seeing this, I'm immediately curious about whether this could somehow be viewed as a power series in $(x,y)$ evaluated at $x=y=1$. $\endgroup$ – Semiclassical Jul 24 '14 at 13:21
  • $\begingroup$ @Semiclassical Yes! Thanks. $\endgroup$ – Olivier Oloa Jul 24 '14 at 13:29
  • $\begingroup$ Is that a "Yes, there is!" or a "Yes, I'd love to know that too!" ? $\endgroup$ – Semiclassical Jul 24 '14 at 13:31
  • 1
    $\begingroup$ @Semiclassical It is "Yes, there is!":) $\endgroup$ – Olivier Oloa Jul 24 '14 at 13:32
  • $\begingroup$ I think it comes off as "Hey, do this for me!" rather than "Hey, try this!" It might help to emphasize that this is a challenge problem. (Also, pointing out that it is listed on the Mathworld page is great, but I think it'd be even better to point out that it's listed explicitly on there as being from you.) $\endgroup$ – Semiclassical Jul 24 '14 at 13:44
12
$\begingroup$

Summation by parts seems to work fine. For $j=1$ we have: $$\sum_{i=1}^{+\infty}\frac{H_{i+1}}{i(i+1)}=2,$$ since $\sum_{i=1}^{N}\frac{1}{i(i+1)}=1-\frac{1}{1+N}$ and so: $$\begin{eqnarray*}\sum_{i=1}^{N}\frac{H_{i+1}}{i(i+1)}&=&H_{N+1}\left(1-\frac{1}{N+1}\right)-\sum_{i=1}^{N-1}\left(1-\frac{1}{i+1}\right)\frac{1}{i+2}\\&=&H_2+\sum_{i=1}^{N-1}\frac{1}{(i+1)(i+2)}+O\left(\frac{\log N}{N}\right)\\&=&H_2+\frac{1}{2}+O\left(\frac{\log N}{N}\right),\end{eqnarray*}$$ since: $$\sum_{i=1}^{+\infty}\frac{1}{(i+1)\cdot\ldots\cdot(i+k)}=\frac{1}{(k-1)k!}.$$ For $j=2$ we have: $$\begin{eqnarray*}1!\sum_{i=1}^{N}\frac{H_{i+2}}{i(i+1)(i+2)}&=&H_{N+1}\left(\frac{1}{4}-\frac{1}{2(N+1)(N+2)}\right)-\sum_{i=1}^{N-1}\left(\frac{1}{4}-\frac{1}{2(i+1)(i+2)}\right)\frac{1}{i+3}\\&=&\frac{H_3}{4}+\frac{1}{4\cdot 3!}+O\left(\frac{\log N}{N}\right)\end{eqnarray*}$$ and in the general case we have: $$(j-1)!\sum_{i=1}^{+\infty}\frac{H_{i+j}}{i\cdot\ldots\cdot(i+j)}=(j-1)!\left(\frac{H_{j+1}}{j^2(j-1)!}+\frac{1}{j^2(j+1)!}\right)=\frac{H_{j+1}}{j^2}+\frac{1}{j^3(j+1)},$$ so the original sum equals: $$\sum_{j=1}^{+\infty}\frac{H_{j+1}}{j^2}+\sum_{j=1}^{+\infty}\frac{1}{j^3(j+1)}=\sum_{j=1}^{+\infty}\frac{H_j}{j^2}+\sum_{j=1}^{+\infty}\frac{1}{j^3}=2\zeta(3)+\zeta(3)=3\zeta(3).$$

$\endgroup$
  • 1
    $\begingroup$ Or as one of my old profs liked to call it, "summagration by parts". I hardly ever think to use this technique so +1. $\endgroup$ – David H Jul 25 '14 at 4:49
6
$\begingroup$

This is another path.

Theorem. Let $x$ be a real number such that $-1<x<1$ and set $$ Z(x):=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{(i-1)! (j-1)!}{(i+j)!} x^{i+j}. $$ Then $$ Z(x)= 2 \:\mathrm{Li_2} (x) - \mathrm{Li_2} \left( x(2-x) \right) \tag1 $$

where $\mathrm{Li_2}$ is the dilogarithm function such that $\mathrm{Li_2}(x)=\displaystyle \sum_{n=1}^{\infty}\displaystyle \frac{x^{n}}{n^2}= - \displaystyle \int_{0}^{x}\,\displaystyle \frac{\log(1-t)}{t}\mathrm{d}t$.

Proof. Observe that $$ {\frac{(i-1)! (j-1)!}{(i+j)!}= \frac{\Gamma(i) \Gamma(j)}{(i+j) \Gamma(i+j)} = \frac{1}{i+j}\,B(i,j)= \int_{0}^{1}\, \frac{t^{i-1}(1-t)^{j-1}}{i+j}\mathrm{d}t.} $$ Let $|x|<1$. We can obtain $$ \begin{align*} \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{(i-1)! (j-1)!}{(i+j)!} x^{i+j} &= \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\displaystyle \int_{0}^{1} \frac{t^{i-1}(1-t)^{j-1}}{i+j}dt \:x^{i+j} \\ &= x^2 \int_{0}^{1} \sum_{j=1}^{\infty}\,(x(1-t))^{j-1}\sum_{i=1}^{\infty} \frac{(tx)^{i-1}}{i+j}\:\mathrm{d}t \\ &= x^2 \int_{0}^{1} \sum_{j=1}^{\infty}\,(x(1-t))^{j-1}(tx)^{-1-j}\sum_{i=j+1}^{\infty} \frac{(tx)^{i}}{i}\:\mathrm{d}t\\ &= x^2 \int_{0}^{1} \sum_{j=1}^{\infty}\,(x(1-t))^{j-1}(tx)^{-1-j}\int_{0}^{tx} \frac{u^{j}}{1-u}\:\mathrm{d}u\:\mathrm{d}t \\ &= -x \int_{0}^{1} \frac{\log(1-x + x u)}{1-x u} \:\mathrm{d}u \\ &= -x \int_{1/x - 1}^{1/x} \frac{ \log(1-\frac{x}{2-x }u) + \log(\frac{2-x}{x})}{u} \:\mathrm{d}u\\ &= 2 \: \mathrm{Li_2} (x) - \mathrm{Li_2} \left( x(2-x) \right) \end{align*}$$

Example 1. $$ \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{(i-1)! (j-1)!}{(i+j)!}= \frac{\pi^2}{6} $$ Put $x=1$ in $(1)$.

Example 2. $$ \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{(i-1)! (j-1)!}{(i+j)!} \frac{1}{ \varphi^{2(i+j)}}= \frac{\pi^2}{30}-\ln^2 \varphi $$ Put $x=1/\varphi^2$ in $(1)$, with $\varphi=\frac{1+\sqrt{5}}{2}$.

Proposition. $$ \sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \displaystyle \frac{(i-1)! (j-1)!}{(i+j)!} H_{i+j} = 3 \: \zeta(3) \tag2 $$

Proof. From the identity $$ 1+x+\cdots+x^{i+j-1} = \frac{1- x^{i+j}}{1 - x\,}, \qquad i\geq1,\,j\geq1, 0<x<1, $$ we write $$ H_{i+j} = \displaystyle \int_{0}^{1} \displaystyle \frac{1- x^{i+j}}{1 - x\,}\: \mathrm{d}x. $$ Then, using $(1)$, we obtain $$ \begin{align} \sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \frac{(i-1)! (j-1)!}{(i+j)!} \: H_{i+j} & =\int_{0}^{1} \frac{ Z(1) - Z(x)}{1 - x}\: \mathrm{d}x \\ & = \int_{0}^{1} \ln(1-x)\left(Z(1) - 2 \: \mathrm{Li_2} (x) + \mathrm{Li_2} \left( x(2-x) \right) \right)'\: \mathrm{d}x \\& = 2 \int_{0}^{1} \dfrac{\ln^2 (1-x)}{2-x}\: \mathrm{d}x \\ & = 2 \int_{0}^{1} \dfrac{\ln^2 x}{1+x}\: \mathrm{d}x \nonumber \\ & = 2 \int_{0}^{1} \sum_{n=0}^{\infty}(-1)^{n} x^{n} \ln^2 x\: \mathrm{d}x \\ & = 3 \: \zeta(3). \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.