question
Compute the limit $$\displaystyle{\lim_{n\to+\infty}a^n(n-\zeta(2)-\zeta(3)-\cdots-\zeta(n))}$$, if any, for the various values of the positive real a, where $\zeta$ the zeta function of Mr. Riemann.
Be areal with $|a|<2$. Prove that $$\displaystyle{\sum_{n\geq2}a^n(n-\zeta(2)-\zeta(3)-\cdots-\zeta(n))=a\left(\frac{\Psi(2- a)+\gamma}{1-a}-1\right)}$$, where $\Psi$ the function digamma.
My work
$$\sum\limits_{k=2}^n \zeta(k) = \sum\limits_{k=2}^n \sum\limits_{m=1}^\infty \frac{1}{m^k} = \sum\limits_{m=1}^\infty \sum\limits_{k=2}^n \frac{1}{m^k} = \sum\limits_{m=1}^\infty \sum\limits_{k=0}^{n-2} \frac{1}{m^{k+2}} = \sum\limits_{m=1}^\infty \frac{1}{m^2} \sum\limits_{k=0}^{n-2} \frac{1}{m^k} = \sum\limits_{m=1}^1 \frac{1}{m^2} \sum\limits_{k=0}^{n-2} \frac{1}{m^k} + \sum\limits_{m=2}^\infty \frac{1}{m^2} \sum\limits_{k=0}^{n-2} \frac{1}{m^k} =$$
$$= n-1 + \sum\limits_{m=2}^\infty \left(\frac{1}{m^2} \cdot \frac{1 - \frac{1}{m^{n-1}}}{1 - \frac{1}{m}}\right) = n-1 + \sum\limits_{m=2}^\infty \frac{m^{n-1} - 1}{(m - 1)m^n} = n-1 + \sum\limits_{m=1}^\infty \frac{1}{(m+1)m} - \sum\limits_{m=2}^\infty \frac{1}{(m-1)m^n} = n - \sum\limits_{m=1}^\infty \frac{1}{m(m+1)^n}$$
So $$a^n \left(n - \sum\limits_{k=2}^n \zeta(k)\right) = a^n \sum\limits_{m=2}^\infty \frac{1}{(m-1)m^n} = \sum\limits_{m=2}^\infty \frac{1}{m-1} \left(\frac{a}{m}\right)^n$$