Infinite sum involving complete and incomplete zeta function

Question

While playing around with complete and incomplete zeta functions I found a nice formula which to my knowledge was not discussed in MSE.

Here it is

$$\sum_{k=2}^\infty (\zeta(k)-1) = 1\tag{1}$$

and its generalizaton

$$\sum _{k=2}^{\infty }\left( \zeta (k)-\left(1+\frac{1}{2^k}+\frac{1}{3^k}+...+\frac{1}{m^k}\right)\right)=\sum_{k=2}^{\infty} (\zeta(k)-H_m^{(k)}) = \frac{1}{m}\tag{2}$$

Here Riemann's zeta function is defined as

$$\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$$

and the incomplete zeta function, which is traditionally called generalized Harmonic number, is defined as

$$H_m^{(k)} = \sum_{n=1}^{m} n^{-k}$$

A corollary of (2) is the interesting relation

$$H_n=\sum _{k=2}^{\infty } \left(n \;\zeta (k)-(n+1) H_n^{(k)}+H_n^{(k-1)}\right)\tag{3}$$

The question asks for a proof of (1), (2), and (3).

Related

A double sum involving the Riemann zeta function

Answer 1

\begin{eqnarray*} \sum_{k=2}^{\infty} ( \zeta(k)-1) =\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{n^k} =\sum_{n=2}^{\infty} \sum_{k=2}^{\infty}\frac{1}{n^k} =\sum_{n=2}^{\infty} \frac{1}{n^2} \frac{1}{1-1/n}=\sum_{n=2}^{\infty} \frac{1}{n(n-1)} =1 \end{eqnarray*} \begin{eqnarray*} \sum_{k=2}^{\infty} ( \zeta(k)-H_m^{(k)}) =\sum_{k=2}^{\infty} \sum_{n=m+1}^{\infty} \frac{1}{n^k} =\sum_{n=m+1}^{\infty} \sum_{k=2}^{\infty}\frac{1}{n^k} =\sum_{n=m+1}^{\infty} \frac{1}{n^2} \frac{1}{1-1/n}=\sum_{n=m+1}^{\infty} \frac{1}{n(n-1)} =\frac{1}{m}. \end{eqnarray*} First equality : Definition of the zeta function.

Second equality : Invert the order of the sums.

Third equality : Geometric sum.

Fourth equality : Partial fractions & telescoping sum.