Series of the day

The sum $\displaystyle \sum\frac{1}{(n+1)^pn^q}$

Today we prove the following result involving the Riemann zeta function $$\sum_{n=0}^{\infty} \frac{ 1}{(n+1)^{r+1}n^{m+1}} = (-1)^m\left[\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) - \sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) + \binom{m+r+1}{r}\right]$$ In the proof requires we use some properties of the digamma and the polygamma functions.

Proof:

The difference between two digamma functions may be expressed as $$(a-b)\sum_{j=0}^{\infty} \frac{1}{(j+a)(j+b)} = \psi(a)-\psi(b)\tag{1}$$ if we make change the of variable $j=n-1$ $$\sum_{n=1}^{\infty} \frac{1}{(n-1+a)(n-1+b)} = \frac{\psi(a)-\psi(b)}{a-b}$$ Recall the Leibnitz rule for higher-order derivatives: $$\frac{d^n}{dx^n} f(x)g(x) = \sum_{j=1}^n \binom{n}{j} f^{(n-j)}(x)g^{(n)}(x)$$ Differentiating $(1)$ both sides with respect to $a$, $r$ times: $$\begin{eqnarray*}\sum_{n=1}^{\infty} \frac{(-1)^r r!}{(n-1+a)^{r+1}(n-1+b)} &=& \sum_{j=0}^{r}\binom{r}{j}\left(\frac{d^{n-j}}{dx^{n-j}} \frac{1}{(a-b)}\right) \left(\frac{d^{j}}{da^{j} } \psi(a) - \psi(b)\right)\\ \\ & = &\sum_{j=0}^{r}\binom{r}{j} \left(\frac{(-1)^{r-j} (r-j) !}{(a-b)^{r-j+1}}\right)\left(\frac{d^{j}}{da^{j} } \psi(a) - \psi(b)\right)\\ &=& \sum_{j=1}^{r}\binom{r}{j}\frac{(-1)^{r-j} (r-j) !\psi^{(j)}(a)}{(a-b)^{r-j+1}} + (-1)^{r} r!\frac{\psi(a)-\psi(b)}{(a-b)^{r+1}} \end{eqnarray*}$$ Where $\psi^{(n)}(x)$ is the polygamma function. This function has the following expansion: $$\psi^{(n)}(x) = n! \sum_{j=0}^{\infty} \left(\frac{-1}{j+x}\right)^{n+1} \tag{2}$$ from this expansion we can derive the following relation: $$\psi^{(n)}(1) = n! (-1)^{n+1}\zeta(n+1) \tag{3}$$ This function also satisfies the following recurrence formula: $$\psi^{(n)}(x+1) =\psi^{(n)}(x) - \frac{n!}{(-x)^{n+1}} \tag{4}$$ So, if we let $a\to 2$ then $\psi(2) =1 -\gamma$: $$\begin{eqnarray*}\sum_{n=1}^{\infty} \frac{ (-1)^rr!}{(n+1)^{r+1}(n-1+b)} & =& \sum_{j=1}^{r}\binom{r}{j}\frac{(-1)^{r-j} (r-j) !\psi^{(j)}(2)}{(2-b)^{r-j+1}} + (-1)^{r} r!\frac{1-\gamma-\psi(b)}{(a-b)^{r+1}}\\ & =& \sum_{j=1}^{r}\binom{r}{j}\frac{(-1)^{r-j} (r-j) !\psi^{(j)}(1)}{(2-b)^{r-j+1}} -\sum_{j=1}^{r}\binom{r}{j}\frac{(-1)^{r-j} (r-j) !j!}{(-1)^{j+1}(2-b)^{r-j+1}} + (-1)^{r} r!\frac{1-\gamma-\psi(b)}{(a-b)^{r+1}}\quad \textrm{ from } (4)\\ \end{eqnarray*}$$ After rearranging the expression: $$\begin{eqnarray*} \sum_{n=1}^{\infty} \frac{ 1}{(n+1)^{r+1}(n-1+b)} &= & -\sum_{j=1}^{r}\frac{ \zeta(j+1)}{(2-b)^{r-j+1}} +\sum_{j=1}^{r}\frac{ 1}{(2-b)^{r-j+1}}+ \frac{1-\gamma-\psi(b)}{(2-b)^{r+1}} \quad \textrm{ from } (3) \end{eqnarray*}$$ If we now differentiate both sides with respect to $b$, $m$ times: $$\begin{eqnarray*}\sum_{n=1}^{\infty} \frac{ (-1)^m m!}{(n+1)^{r+1}(n-1+b)^{m+1}} & = & -\sum_{j=1}^{r}\frac{ (r-j+m)!\zeta(j+1)}{(r-j)!(2-b)^{r-j+m+2}} + \sum_{j=1}^{r}\frac{ (r-j+m)!}{(r-j)!(2-b)^{r-j+1}} \\ &-& \sum_{k=1}^{m}\binom{m}{k}\frac{ (r+m-k) !\psi^{k}(b)}{r! (2-b)^{r+m-k+2}} + \frac{(r+m)!}{r!} \frac{(1-\gamma - \psi(b))}{(2-b)^{r+m+2}}\\ \end{eqnarray*}$$ If we let $b\to 1$ then $\psi(1) = -\gamma$ and using again the relationship between $\psi^{(n)}(1)$ and $\zeta(n+1)$ $$\begin{eqnarray*}\sum_{n=0}^{\infty} \frac{ 1}{(n+1)^{r+1}n^{m+1}} & = & -(-1)^m\sum_{j=1}^{r}\frac{ (r-j+m)!}{(r-j)!m!} \zeta(j+1) + (-1)^m\sum_{j=1}^{r}\frac{ (r-j+m)!}{(r-j)!m!}\\ &+& (-1)^m\sum_{k=1}^{m}\frac{(r+m-k) !}{(m-k)!r!}(-1)^k\zeta(k+1) + (-1)^m \frac{(r+m)!}{r! m!} \\ & = & -(-1)^m\sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) + (-1)^m\sum_{j=1}^{r}\binom{ r-j+m}{m}\\ &+& (-1)^m\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) + (-1)^m\binom{r+m}{m} \end{eqnarray*}$$ We can prove by induction that $$\sum_{j=1}^{r}\binom{ r-j+m}{m} = \binom{r+m}{m+1}$$ Base case, $r=1$ $$\sum_{j=1}^{1}\binom{ 1-j+m}{m} = \binom{ m}{m} = \binom{ m+1}{m+1}$$ Inductive step, suppose the formula is valid for $r = p$ $$\sum_{j=1}^{p}\binom{ p-j+m}{m} = \binom{p+m}{m+1}$$ Then, for $r=p+1$: $$\begin{eqnarray*} \sum_{j=1}^{p+1}\binom{ p+1-j+m}{m} & =& \sum_{k=0}^{p}\binom{ p-k+m}{m} \\ & =& \binom{p+m}{m} + \sum_{k=1}^{p}\binom{ p-k+m}{m}\\ & = & \binom{p+m}{m} + \binom{p+m}{m+1}\\ & =& \binom{p+1+m}{m+1 } \end{eqnarray*}$$ Therefore: $$\begin{eqnarray*} \sum_{n=0}^{\infty} \frac{ 1}{(n+1)^{r+1}n^{m+1}} &=& (-1)^m\left[\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) - \sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) +\binom{r+m}{m+1} + \binom{r+m}{m} \right] \\ &=& (-1)^m\left[\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) - \sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) +\binom{r+m+1}{r} \right] \end{eqnarray*}$$ Hence $$\boxed{\sum_{n=0}^{\infty} \frac{ 1}{(n+1)^{r+1}n^{m+1}} = (-1)^m\left[\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) - \sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) + \binom{m+r+1}{r}\right]}$$