Decomposition of $\psi^{(n)}(1)$ in terms of $\psi^{(n)}(k)$ 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?
 A: 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
$$
\eqalign{
  & \sum\limits_{k = 2}^\infty  {\left( { - 1} \right)^{\,k} \psi ^{\,\left( n \right)} (k)}  = \sum\limits_{1\, \le \,j\,} {\left( {\psi ^{\,\left( n \right)} (2j) - \psi ^{\,\left( n \right)} (2j + 1)} \right)}  =   \cr 
  &  =  - \sum\limits_{1\, \le \,j\,} {\left. {\Delta _{\,z} \,\psi ^{\,\left( n \right)} (z)} \right|_{z = 2j} }  =  - \left( { - 1} \right)^{\,n} n!\sum\limits_{1\, \le \,j\,} {\;\left( {2j} \right)^{\, - n - 1} }  =   \cr 
  &  = 2^{\, - n - 1} \left( { - 1} \right)^{\,n + 1} n!\sum\limits_{1\, \le \,j\,} {\;{1 \over {j^{\,n + 1} }}}
  = 2^{\, - n - 1} \left( { - 1} \right)^{\,n + 1} n!\sum\limits_{0\, \le \,k\,} {\;{1 \over {\left( {k + 1} \right)^{\,n + 1} }}}  =   \cr 
  &  = 2^{\, - n - 1} \psi ^{\,\left( n \right)} (1) \cr} 
$$
A: 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}.
