In this question it is shown, up to a small detail or two, that $$\lim_{n\to\infty}n\cdot \sum_{m=1}^{\infty}\Big(1-\frac{1}{m}\Big)^n\cdot \frac{1}{m^2}=1$$ The proof essentially involves Taylor series and reimagining the sum as a Riemann integral to show a lower bound and an upper bound each converge to $1$.
However, I believe the problem can be solved a different way. If we use the binomial theorem to expand $(1-1/m)^n$ and swap the resulting double sum, we have $$ \lim_{n\to\infty}n\sum_{k=0}^n \binom{n}{k}(-1)^k\zeta (k+2) $$Intuitively, this limit should converge because $\zeta(n+2)\to 1$ as $n\to\infty$, and then the series looks like $\sum_{0\le k\le n}\binom{n}{k}(-1)^k=0$, but the asymptotics evade me. Previously I was interested in so-called 'alternating binomial zeta series,' (my own clunky descriptor of this object) and found that convergence was quite delicate and subtle. I tried using Stolz-Cesaro, the discrete version of L'Hopital's Rule, but was unsusccessful, though maybe it's possible through this or other means.
Not an answer, but an interesting formula.
Let $p_j=\zeta(j)-1$ for $j>1.$
Let: $$A_{n,i}=\sum_{k=0}^{n}(-1)^k\binom nk p_{k+i},$$ for $i>1,$ then we get $$\begin{align}A_{n+1,i}&=\sum_{k=0}^{n+1}(-1)^k\left(\binom{n}{k}+\binom{n}{k-1} \right)p_{k+i}\\&=A_{n,i} -A_{n,i+1}\tag{1}\end{align}$$
So if $B_{n,i}=(n+i)A_{n,i},$ then $$B_{n+1,i}=\left(B_{n,i}-B_{n,i+1}\right) + A_{n,i}$$
Now you are trying to prove $B_{n,2}\to 1.$ (Technically, you want $\frac{n}{n+2}B_{n,2}\to 1,$ but that is the same thing.)
Substituting again, we get:
$$\begin{align}B_{n+2,i}&=\left(\left(B_{n,i}-B_{n,i+1}\right)+A_{n,i}\right)-\left((B_{n,i+1}-B_{n,i+2})+A_{n,i+1}\right)+A_{n+1,i}\\ &=\left(B_{n,i}-2B_{n,i+1}+B_{n,i+2}\right)+A_{n,i}+A_{n+1,i}-A_{n,i+1}\\ &=2A_{n+1,i} \end{align}$$ The last step by (1).
Inductively you'll get:
$$B_{n+m,i}=\sum_{j=0}^m(-1)^m \binom{m}{j}B_{n,i+j}+mA_{n+m-1,i} $$
Using $B_{1,i}=A_{1,i}=p_{i}-p_{i+1},$ we get: $$B_{n,2}=\sum_{j=0}^{n-1}(-1)^{j}\binom{n-1}{j}(p_{2+j}-p_{3+j}) +B_{n-1,2}$$
So: $$\begin{align}B_{n,2}&=\sum_{k=0}^{n-1} \sum_{j=0}^{k}(-1)^{j}\binom{k}{j}(p_{2+j}-p_{3+j})\\&=\sum_{j=0}^{n-1}(-1)^j(p_{2+j}-p_{3+j})\sum_{k=j}^{n-1}\binom{k}{j}\\&=\sum_{j=0}^{n-1}(-1)^j\binom{n}{j+1}(p_{2+j}-p_{3+j})\\ &=\sum_{j=0}^{n-1}(-1)^{j}\binom n{j+1}p_{2+j}+\sum_{j=1}^{n}(-1)^{j}\binom n{j}p_{2+j}\\ &=np_{2}+(-1)^np_{2+n}+\sum_{j=1}^{n-1}(-1)^{j}\left(\binom n{j+1}+\binom nj\right)p_{2+j}\\ &=np_{2}+(-1)^np_{2+n}+\sum_{j=1}^{n-1}(-1)^j\binom {n+1}{j+1}p_{2+j}\\ &=-p_{2}+\sum_{k=1}^{n+1}(-1)^{k-1}\binom{n+1}{k}p_{1+k} \end{align}$$
(This section is broken.)
We can try generating functions:
If $$P_i(z)=\sum_{n=0}^{\infty}\frac{p_{n+i}}{n!}z^n$$
Writing:
$$\begin{align}P_i(z)&=\sum_{k=2}^{\infty}\sum_{n=0}^{\infty}\frac{1}{n!}\frac{z^n}{k^{n+i}}\\&=\sum_{k=2}^{\infty}\frac{e^{z/n}}{n^i}\end{align}$$
We also get: $$A_i(z)=\sum_{n=0}^{\infty} A_{n,i}z^n/n!=e^zP_i(-z).$$
From this we get $A_i'(z)=A_i(z)-A_{i+1}(z).$ That is equivalent to (1) above.
If $c_{n,i}=nA_{n,i}$ then $$C_i(z)=\sum_{n=1}^{\infty}c_{n,i}z^{n}/n!=zA_i'(z)=ze^z(P_i(z)-P_{i+1}(z)).$$