Decomposition of $\psi^{(n)}(1)$ in terms of $\psi^{(n)}(k)$

Question

Accidentally run into this identity:

\begin{align} \psi^{(n)}(1) &= 2^{n+1}\, \sum_{k = 2}^\infty (-1)^k\,\psi^{(n)}(k) \tag{1}\label{1} , \end{align}

its variation

\begin{align} 2^{-n-1} &= \sum_{k = 1}^\infty (-1)^{k+1}\,\frac{\psi^{(n)}(k+1)}{\psi^{(n)}(1)} \tag{2}\label{2} \end{align}

and related

\begin{align} \psi^{(2m-1)}(1) &=\tfrac1m\,(-4)^{m-1}\,\pi^{2m}\,\operatorname{B}_{2m} \tag{3}\label{3} , \end{align}
where $\operatorname{B}_{2m}$ is $2m$-th Bernoulli number.

WolframAlpha helps to confirm \eqref{1}, \eqref{2} for small values of $n$, but does not recognize it for general $n$.

Question: Is this a well-known set of identities?

Answer 1

It is well known that $$ \eqalign{ & \psi ^{\,\left( n \right)} (z) = {{d^{\,n} } \over {d\,z^{\,n} }}\psi (z)\quad \;\left| \matrix{ \;n \in \;\; \mathbb Z\,_ + \;\;\, \hfill \cr \;0 < {\mathop{\rm Re}\nolimits} (z) \hfill \cr} \right.\quad = \cr & = \left( { - 1} \right)^{\,n + 1} n!\sum\nolimits_{\;j\, = \;0\;}^{\;\infty } {{1 \over {\left( {j + z} \right)^{\,n + 1} }}} \cr} $$

Then $$ \eqalign{ & \Delta _{\,z} \,\psi ^{\,\left( n \right)} (z) = \;\psi ^{\,\left( n \right)} (z + 1) - \psi ^{\,\left( n \right)} (z) = \cr & = \left( { - 1} \right)^{\,n + 1} n!\left( {\sum\nolimits_{\;j\, = \;0\;}^{\;\infty } {{1 \over {\left( {j + z + 1} \right)^{\,n + 1} }} - {1 \over {\left( {j + z} \right)^{\,n + 1} }}} } \right) = \cr & \left( { - 1} \right)^{\,n} n!\;z^{\, - n - 1} \cr} $$ and from that $$ \eqalign{ & \sum\limits_{k = 2}^\infty {\left( { - 1} \right)^{\,k} \psi ^{\,\left( n \right)} (k)} = \sum\limits_{1\, \le \,j\,} {\left( {\psi ^{\,\left( n \right)} (2j) - \psi ^{\,\left( n \right)} (2j + 1)} \right)} = \cr & = - \sum\limits_{1\, \le \,j\,} {\left. {\Delta _{\,z} \,\psi ^{\,\left( n \right)} (z)} \right|_{z = 2j} } = - \left( { - 1} \right)^{\,n} n!\sum\limits_{1\, \le \,j\,} {\;\left( {2j} \right)^{\, - n - 1} } = \cr & = 2^{\, - n - 1} \left( { - 1} \right)^{\,n + 1} n!\sum\limits_{1\, \le \,j\,} {\;{1 \over {j^{\,n + 1} }}} = 2^{\, - n - 1} \left( { - 1} \right)^{\,n + 1} n!\sum\limits_{0\, \le \,k\,} {\;{1 \over {\left( {k + 1} \right)^{\,n + 1} }}} = \cr & = 2^{\, - n - 1} \psi ^{\,\left( n \right)} (1) \cr} $$

Answer 2

Consider the first identity

\begin{align} \psi^{(n)}(1) &= 2^{n+1}\,\sum_{k = 2}^\infty(-1)^k\,\psi^{(n)}(k) \tag{1}\label{1} \end{align}

From this paper

Batir, N., 2007. On some properties of digamma and polygamma functions. Journal of Mathematical Analysis and Applications, 328(1), pp.452-465.

\begin{align} \psi^{(n)}(x)&=(-1)^{n+1}\int_0^\infty\frac{t^n\,\exp(-xt)}{1-\exp(-t)}\, dt \tag{2}\label{2} \\ &=(-1)^{n+1}n!\sum_{i=0}^\infty \frac 1{(x+k)^{n+1}} \tag{3}\label{3} , \end{align}

so we can rewrite \eqref{1} as

\begin{align} \psi^{(n)}(1) &= (-2)^{n+1}\, \sum_{k = 2}^\infty (-1)^{k}\, \int_0^\infty \frac{t^n\,\exp(-kt)}{1-\exp(-t)}\, dt \tag{4}\label{4} \\ &= (-2)^{n+1}\, \int_0^\infty \left( \sum_{k = 2}^\infty (-1)^{k}\, \frac{t^n\,\exp(-kt)}{1-\exp(-t)} \right) \, dt \tag{5}\label{5} \\ &= (-2)^{n+1}\, \int_0^\infty \frac{t^n}{1-\exp(-t)} \left( \sum_{k = 2}^\infty (-1)^{k}\, \exp(-kt) \right) \, dt \tag{6}\label{6} \end{align}

\begin{align} &= (-2)^{n+1}\, \int_0^\infty \frac{t^n}{1-\exp(-t)} \left( \frac{\exp(-t)}{1+\exp(t)} \right) \, dt \tag{7}\label{7} \\ &= (-2)^{n+1}\, \int_0^\infty \frac{t^n}{\exp(2t)-1} \, dt \tag{8}\label{8} \\ &= 2\, \int_0^\infty \frac{(-2t)^n}{1-\exp(2t)} \, dt \overset{\color{blue}{x=\exp(-2t)}}{=} \int_0^1 \frac{\ln(x)^n}{x-1}\, dx \tag{9}\label{9} . \end{align}

The last integral is known, \begin{align} \int_0^1 \frac{\ln(x)^n}{x-1}\, dx &= (-1)^{n+1}n!\zeta(n+1) =\psi^{(n)}(1) \tag{10}\label{10} , \end{align}

so we have \eqref{1}.