Help on how to show that $3-\zeta(2)=\sum_{n=3}^{\infty}n[\zeta(n)-1]?$ How do we go about showing that

$$3-\zeta(2)=\sum_{n=3}^{\infty}n[\zeta(n)-1]?$$

 A: Due to normal convergence, one is allowed to write
$$
\begin{align}
\sum_{n=3}^{\infty}n[\zeta(n)-1]&=\sum_{n=3}^{\infty}\left(n\sum_{k=2}^{\infty}\frac1{k^n}\right)
\\\\&=\sum_{k=2}^{\infty}\left(\sum_{n=3}^{\infty}n\frac1{k^{n}}\right)
\\\\&=\sum_{k=2}^{\infty}\frac{3k-2}{k^2(k-1)^2}
\\\\&=\sum_{k=2}^{\infty}\frac{1}{(k-1)^2}-\sum_{k=2}^{\infty}\frac{2}{k^2}-\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{k-1}\right)
\\\\&=-\frac{\pi ^2}{6}+3
\end{align}
$$ where we have used a standard result and a telescoping series.
A: We can also use the power series $$\sum_{k=2}^{\infty}\zeta\left(k\right)x^{k-1}=-\psi^{\left(0\right)}\left(1-x\right)-\gamma,\,\left|x\right|<1
 $$ where $\psi^{\left(0\right)}\left(1-x\right)
 $ is the Digamma function. So we have $$\sum_{k=3}^{\infty}x^{k-1}\left(\zeta\left(k\right)-1\right)=-\psi^{\left(0\right)}\left(1-x\right)-\gamma-x\zeta\left(2\right)+\frac{x^{2}}{x-1}
 $$ then if we multiply both by $x$ we get $$\sum_{k=3}^{\infty}x^{k}\left(\zeta\left(k\right)-1\right)=-x\psi\left(1-x\right)-x\gamma-x^{2}\zeta\left(2\right)+\frac{x^{3}}{x-1}
 $$ and now if we differentiate we obtain $$\sum_{k=3}^{\infty}kx^{k-1}\left(\zeta\left(k\right)-1\right)=x\psi^{\left(1\right)}\left(1-x\right)-\psi^{\left(0\right)}\left(1-x\right)-\gamma-2x\zeta\left(2\right)+\frac{x^{2}\left(2x-3\right)}{\left(x-1\right)^{2}}
 $$ where $\psi^{\left(1\right)}\left(1-x\right)
 $ is the Trigamma function. Now since $$\psi^{\left(n\right)}\left(1-z\right)+\left(-1\right)^{n-1}\psi^{\left(n\right)}\left(z\right)=\pi\frac{d^{n}}{dx^{n}}\cot\left(\pi x\right)
 $$ and $$\lim_{x\rightarrow1}{\pi\left(\cot\left(\pi x\right)-\pi\csc^{2}\left(\pi x\right)\right)}{\left(1-x\right)^{2}}=-1
 $$ and we know the value $$\psi^{\left(0\right)}\left(1\right)=-\gamma,\,\psi^{\left(1\right)}\left(1\right)=\zeta\left(2\right)
 $$ we finally have $$\sum_{k=3}^{\infty}k\left(\zeta\left(k\right)-1\right)=\color{red}{3-\zeta\left(2\right)}$$ as wanted.
A: $$\sum_{n\geq 2}\left(\zeta(n)-1\right)x^n = \sum_{n\geq 2}\sum_{m\geq 2}\frac{x^n}{m^n} =\sum_{m\geq 2}\frac{x^2}{m(m-x)}\tag{1}$$
by differentiation leads to:
$$ \sum_{n\geq 2}\left(\zeta(n)-1\right)n x^{n-1} = \sum_{m\geq 2}\left(-\frac{1}{m}+\frac{m}{(m-x)^2}\right)\tag{2} $$
and by evaluating both sides at $x=1$ we get:
$$ \sum_{n\geq 2}n\left(\zeta(n)-1\right) = \sum_{h\geq 1}\left(\frac{1}{h}-\frac{1}{h+1}+\frac{1}{h^2}\right)=1+\zeta(2).\tag{3} $$
