How are you going to solve this?

Let $f(n) = n - \sum_{k=2}^{n} \zeta(n)$ . Then, $\begin{array}{lll} f(n) & =& 1-\displaystyle \sum_{k=2}^{n} \left(\zeta(k)-1\right)\\ & = & 1-\displaystyle\sum_{k=2}^{n} \displaystyle\sum_{m=2}^{\infty} \dfrac{1}{m^k}\\ & = & 1 - \displaystyle\sum_{m=2}^{\infty} \displaystyle\sum_{k=2}^{n}\dfrac{1}{m^k}\\ & = & 1 - \displaystyle\sum_{m=2}^{\infty} \left(\dfrac{1}{m(m-1)} - \dfrac{1}{m^n(m-1)}\right)\\ & = & \displaystyle\sum_{m=2}^{\infty} \dfrac{1}{m^n(m-1)}\end{array}$

Now I would like to show that $\lim_{n\to\infty} 2^n f(n) =1$ . To do this, observe that for all $n\ge 1$ and $m\ge 3$ we have $\left(\frac{m}{2}\right)^n >\frac{(m-1)n}{2}$ (which can be proved by induction). Therefore,

$\begin{array}{lll} 2^n f(n) &=& \displaystyle\sum_{m=2}^{\infty} \dfrac{1}{(m-1)}\left(\dfrac{2}{m}\right)^n\\ & < & 1+ \displaystyle\sum_{m=3}^{\infty} \dfrac{2} {(m-1)^2 n}\\ 2^n f(n) & < & 1 + \dfrac{2(\zeta(2)-1)}{n}\end{array}$

Thus, by the squeeze theorem, $\lim_{n\to\infty} 2^n f(n) = 1$ .

Therefore, $\lim_{n\to\infty} \dfrac{f(n)}{f(n-1)}=\dfrac{1}{2}\lim_{n\to\infty} \dfrac{2^n f(n)}{2^{n-1} f(n-1)} = \boxed{\dfrac{1}{2}}$

I think the answer of $\dfrac{1}{2}$ is correct for this, not the way you have it posted

$\displaystyle \lim _{ n\rightarrow \infty }{ \dfrac { n-\zeta (2)-\zeta (3)...-\zeta (n+1) }{ n-1-\zeta (2)-\zeta (3)-...-\zeta (n+1) } }$

Then, using L'Hospital, we have

$\displaystyle \lim _{ n\rightarrow \infty }{ \dfrac { n- (n+1) }{ n-1- (n+1) } }=\dfrac{1}{2}$