# For $n, m \ge 0, n+m\ge 2$, evaluate $W_{m, n} =\sum_{k=1}^{\infty} \frac{1}{(k+1)^mk^{n}}$.

For $n, m \ge 0, n+m\ge 2$, what is the value of $W_{m, n} =\sum_{k=1}^{\infty} \frac{1}{(k+1)^mk^{n}}$.

My results:

• From the definition \begin{align*} W_{0, n} &= \zeta(n) \\ W_{m, 0} &= \zeta(m)-1 \text{.} \end{align*}

• Recurrence from which everything else follows: $$W_{m, n} = W_{m-1, n}-W_{m, n-1} \text{.}$$

• Explicit values: \begin{align*} W_{1, n} &= \sum_{k=0}^{n-2} (-1)^k\zeta(n-k) + (-1)^{n-1} \\ W_{m, 1} &=-\sum_{k=0}^{n-2} \zeta(n-k) + m \text{.} \end{align*}

I have not yet worked out the general formula for $W_{m,n}$.

• Sorry to be pedantic: Is there a question here? Is it, "What is the formula for general $W_{m,n}$?" – Eric Towers May 18 '18 at 2:54
• I have a related question at math.stackexchange.com/questions/2783981/… where I've identified the coeffecients as having a Mellin transform inverse pair, I've also got a paper in progress at github.com/crowlogic/a/blob/master/… there is some already known stuff about it at linas.org/math/gkw.pdf as well. It is derived from the transfer operator of the Lüroth series map which you can read about at the link above – crow May 18 '18 at 2:55
• Yep, that is the same question, aside from the binomial coefficient. I didn't see an explicit formula for the coefficients, though – marty cohen May 18 '18 at 4:55

The eigenvalues are given by $$\begin{array}{cl} \lambda_m & = W_{m, m}\\ & = \sum_{n = 1}^{\infty} \frac{1}{n^{m + 1} (n + 1)^{m + 1}}\\ & = \sum_{n = 0}^{\frac{m}{2} + \frac{1}{4} - \frac{(- 1)^m}{4}} \zeta (2 n) \frac{2 (- 1)^{m + 1 - 2 n} \Gamma (2 m - 2 n + 2)}{\Gamma (m + 2 - 2 n) \Gamma (m + 1)} \end{array}$$ where $$\frac{m}{2} + \frac{1}{4} - \frac{(- 1)^m}{4} = [0, 1, 1, 2, 2, 3, 3, 4, 4, 5, \ldots] {where} m = 0, 1, 2, \ldots, 9, \ldots$$ This form of the function means that $Y_{m, n, k} = 0$ when $k$ is odd and $m = n$ so that $$Y_{m, m, k} = \left\{ \begin{array}{ll} 0 & k {is} {odd}\\ \frac{2 (- 1)^{m + 1 - k} \Gamma (2 m - k + 2)}{\Gamma (m + 2 - k) \Gamma (m + 1)} & k {is} {even} \end{array} \right.$$ when $m \neq n$ the off-diagonals involve also odd-integer values of $\zeta$ so we need to find a general expression for $Y_{m, n, k}$ and not just the diagonals $Y_{m, m, k}$ where

$$\begin{array}{ll} W_{m, n} & = \binom{n}{m} \sum_{k = 1}^{\infty} \frac{1}{(k + 1)^{n + 1} k^{m + 1}}\\ & = \binom{n}{m} \sum_{k = 0}^{\max (m, n) + 1} \zeta (k) Y_{m, n, k} \end{array}$$

also

$$Y_{m, n, 0} = \frac{2 (- 1)^{m + n + 1} \Gamma (m + n + 2)}{(n + 1) \Gamma (m + 1)^2 \Gamma (n + 1 - m)} = - \frac{2}{\Gamma (m)^2} \prod_{k = 1}^{2 n - 1} (- 1)^{n - 1} \left\{ \begin{array}{ll} 1 & k = m\\ n - m + k & k \neq m \end{array} \right. \label{Y0}$$ so that $$\begin{array}{lfl} \hat{y}_{n, 0} (s) & = \sum_{m = 0}^n \frac{Y_{m, n, 0}}{m + s}\\ & = \sum_{m = 0}^n \frac{\frac{2 (- 1)^{m + n + 1} \Gamma (m + n + 2)}{(n + 1) \Gamma (m + 1)^2 \Gamma (n + 1 - m)}}{m + s}\\ & = \mathit{} - 2 \hspace{0.17em} \frac{(- 1)^n_{} 3F2 \left[ \begin{array}{c} \begin{array}{lllll} s & & - n & & n + 2 \end{array}\\ \begin{array}{lll} 1 & & s + 1 \end{array} \end{array} ; \hspace{0.17em} 1 \right]}{s} \end{array}$$

where 3F2 is a generalized hypergeometric function and

$$Y_{m, n, 1} = 0 \label{Y1}$$ and $$Y_{m, n, 2} = 0^{m + n} + \frac{(- 1)^{m + n} \Gamma (n + m)}{\Gamma (m + 1)^2 \Gamma (n - m)} \label{Y2}$$ and the coeffecients of $\zeta (3)$ are simply related to a shifted version of the coeffecients of $\zeta (2)$ $$Y_{m, n, 3} = Y_{m, n - 1, 2} \label{Y3}$$

$\begin{array}{ll} Y_{m, n, 4} & = {Res} (\hat{y}_{n, 4} (s), s = - m)\\ & = \lim_{s = - m} (s + m) \hat{y}_{n, 4} (s)\\ & = {Res} (\hat{y}_{n, 4} (s), s = - m)\\ & = \left\{ \begin{array}{ll} 0 & n \leqslant 2\\ \frac{(- 1)^{m + n + 1}}{(m!^{})^2} \left( \prod_{k = 0}^{m - 1} (n - k) \right) \left( \prod_{k = 1}^{m - 2} (n + k) \right) (n^2 - (m + 1) n + m (m - 1)) & n \geqslant 3 \end{array} \right. \end{array}$

we also have

$Y_{m, n, 5} = \frac{(- 1)^{1 + n - m} \left( m^2 + n^2 - 3 \hspace{0.17em} m + n \right) \prod_{k = 1}^{2 \hspace{0.17em} m - 2} (k + 1 + n - m)}{(m!)^2}$

if we write the polynomials specifying the coefficients as hyper-geometric functions we find

$$\begin{array}{ll} y_{n, 0} (x) & = 2 \hspace{0.17em} (- 1)^{n + 1}_{} 2F1 (0 - n, n + 2 ; \hspace{0.17em} 1 ; \hspace{0.17em} x)\\ y_{n, 1} (x) & = 0\\ y_{n, 2} (x) & = \hspace{0.17em} (- 1)^n_{} 2F1 (n, 1 - n ; \hspace{0.17em} 1 ; \hspace{0.17em} x)\\ y_{n, 3} (x) & = (- 1)^{n + 1} 2F1 (n - 1, 2 - n ; 1 ; x)\\ y_{n, 4} (x) & = (- 1)^n 4F3 \left( \begin{array}{l} \begin{array}{lll} {}[0 - n] & [n - 1] & \left[ \frac{1}{2} - \frac{n}{2} \pm \frac{1}{2} \hspace{0.17em} \sqrt{- 3 \hspace{0.17em} n^2 + 6 \hspace{0.17em} n + 1} \right] \end{array}\\ \begin{array}{ll} {}[1] & \left[ - \frac{1}{2} - \frac{n}{2} \pm \frac{1}{2} \hspace{0.17em} \sqrt{- 3 \hspace{0.17em} n^2 + 6 \hspace{0.17em} n + 1} \right] \end{array} \end{array} ; x \right)\\ y_{n, 5} (x) & = (- 1)^{n + 1}_{} 4F3 \left( \begin{array}{l} \begin{array}{lll} {}[n] & [- n - 1] & \left[ - \frac{1}{2} \pm \frac{1}{2} \hspace{0.17em} \sqrt{- 4 \hspace{0.17em} n^2 - 4 \hspace{0.17em} n + 9} \right] \end{array}\\ \begin{array}{ll} {}[1] & \left[ - \frac{3}{2} \pm \frac{1}{2} \hspace{0.17em} \sqrt{- 4 \hspace{0.17em} n^2 - 4 \hspace{0.17em} n + 9} \right] \end{array} \end{array} ; x \right) \end{array}$$ where $[a \pm b]$ is actually two elements, $[a + b]$ and $[a - b]$

• Very impressive. You have certainly taken this much further than I could. Accepted. – marty cohen May 18 '18 at 5:31
• Thanks. It would be nice to know the entire formula – crow May 18 '18 at 13:43
• You might find the related question at math.stackexchange.com/questions/2787558/… interesting – crow May 19 '18 at 21:41

Just to be complete, here is everything I have got so far, including two recurrences at the end that don't seem to be of much use.

According to Wolfy,

$W_{1,1} =1\\ W_{1,2} =\dfrac{\pi^2}{6}-1\\ W_{2, 1} =2-\dfrac{\pi^2}{6}\\ W_{2, 2} =\dfrac{\pi^2}{3}-3\\ W_{1, 3} =\zeta(3)+1-\dfrac{\pi^2}{6}\\ W_{3, 1} =-\zeta(3)-\dfrac{\pi^2}{6}+3\\ W_{2, 3} =\zeta(3)+4-\dfrac{\pi^2}{2}\\ W_{3,2} =\zeta(3)+\dfrac{\pi^2}{2}-6\\ W_{3, 3} =10-\pi^2\\$

$W_{0, n} =\zeta(n)$.

$W_{m, 0} =\zeta(m)-1$.

$\begin{array}\\ W_{m+1, n+1} &=\sum_{k=1}^{\infty} \frac{1}{(k+1)^{m+1}k^{n+1}}\\ &=\sum_{k=1}^{\infty} \frac{1}{(k+1)k(k+1)^mk^{n}}\\ &=\sum_{k=1}^{\infty} \frac{1}{(k+1)^mk^{n}}(\frac1{k}-\frac1{k+1})\\ &=\sum_{k=1}^{\infty} \frac{1}{(k+1)^mk^{n+1}}-\sum_{k=1}^{\infty} \frac{1}{(k+1)^{m+1}k^{n}}\\ &=W_{m, n+1}-W_{m+1, n}\\ \end{array}$

or $W_{m, n} =W_{m-1, n}-W_{m, n-1}$.

$m=0$:

$W_{1, n+1}\\ =W_{0, n+1}-W_{1, n}\\ =\zeta(n+1)-W_{1, n}\\ =\zeta(n+1)-\zeta(n)+W_{1, n-1}\\ =...\pm\zeta(2)\mp 1\\$

so $W_{1,n} =\sum_{k=0}^{n-2} (-1)^k\zeta(n-k) +(-1)^{n-1}$.

$n=0$:

$W_{m+1, 1}\\ =W_{m, 1}-W_{m+1, 0}\\ =W_{m, 1}-\zeta(m+1)+1\\ =-\zeta(m+1)+1+W_{m, 1}\\ =-\zeta(m+1)-\zeta(m)+2+W_{m-1, 1}\\ =...-\zeta(2)+m-1+W_{1, 1}\\$

so $W_{m, 1} =-\sum_{k=0}^{n-2} \zeta(n-k) +m$.

Check:

$(3, 3)\\ =(2, 3)-(3, 2)\\ =((1, 3)-(2,2))-((2,2)-(3, 1))\\ =(1, 3)-2(2, 2)+(3, 1)\\ =(z(3)-z(2)+1)-2((1, 2)-(2, 1))+(-z(3)-z(2)+3)\\ =-2z(2)+4)-2((1, 2)-(2, 1))\\ =-2z(2)+4)-2(z(2)-1-(-z(2)+2)\\ =-6z(2)+10\\ =-\pi^2+10\\$

$\begin{array}\\ W_{m, n}-\zeta(m+n) &=\sum_{k=1}^{\infty} (\frac{1}{(k+1)^mk^{n}}-\frac1{k^{m+n}})\\ &=\sum_{k=1}^{\infty} \frac1{k^n}(\frac{1}{(k+1)^m}-\frac1{k^{m}})\\ &=-\sum_{k=1}^{\infty} \frac1{k^n}(\frac{(k+1)^m-k^m}{(k+1)^mk^{m}})\\ &=-\sum_{k=1}^{\infty} \frac1{(k+1)^mk^{m+n}}\sum_{j=0}^{m-1}\binom{m}{j}k^j\\ &=-\sum_{j=0}^{m-1}\binom{m}{j}\sum_{k=1}^{\infty} \frac1{(k+1)^mk^{m+n-j}}\\ &=-\sum_{j=0}^{m-1}\binom{m}{j}W_{m, m+n-j}\\ \end{array}$

so, isolating the $j=0$ term,

$W_{m, m+n}\\ =\zeta(m+n)-W_{m, n}-\sum_{j=1}^{m-1}\binom{m}{j}W_{m, m+n-j}\\ =\zeta(m+n)-W_{m, n}-\sum_{j=1}^{m-1}\binom{m}{j}W_{m, n+j}\\ =\zeta(m+n)-\sum_{j=0}^{m-1}\binom{m}{j}W_{m, n+j}$.

Similarly, for $n$

$\begin{array}\\ W_{m, n}-\zeta(n) &=\sum_{k=1}^{\infty} (\frac{1}{(k+1)^mk^{n}}-\frac1{k^{n}})\\ &=\sum_{k=1}^{\infty} \frac1{k^n}(\frac{1}{(k+1)^m}-1)\\ &=-\sum_{k=1}^{\infty} \frac1{k^n}(\frac{(k+1)^m-1}{(k+1)^m})\\ &=-\sum_{k=1}^{\infty} \frac1{(k+1)^mk^{n}}\sum_{j=1}^{m}\binom{m}{j}k^j\\ &=-\sum_{j=1}^{m}\binom{m}{j}\sum_{k=1}^{\infty} \frac1{(k+1)^mk^{n-j}}\\ &=-\sum_{j=1}^{m}\binom{m}{j}W_{m, n-j}\\ \end{array}$

so that $W_{m, n} =\zeta(n)-\sum_{j=1}^{m}\binom{m}{j}W_{m, n-j}$.