Directly showing $\lim_{n\to\infty}n\sum_{k=0}^n \binom{n}{k}(-1)^k\zeta (k+2) =1$ 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.
 A: A proof of the asymptotics for
$$ H_{\nu}(x) := \sum_{k>\text{Re}\,\nu+1} (-1)^k\binom{x}{k} \zeta(n-\nu) $$
appears in the publication 'On Mellin-Barnes Type of Integrals and Sums Associated with the Riemann Zeta-Function,' by M. Katsurada, in Public. de L'Institut Mathematique Nouvell serie, tome 62, (76) 1997, 13-25.
Theorem 4.1 states that, for $\nu$ not in the set {-1, 0, 1, 2, ...}
$$H_{\nu}(x)  = \frac{\Gamma(1+x)\,\Gamma(-v-1)}{\Gamma(x-\nu)} - \sum_{k=0}^{[\text{Re}\,\nu+1]} (-1)^k\binom{x}{k} \zeta(n-\nu) + \cal{R}_\nu(x)$$
The proposer's question is for $\nu = -2.$  The empty sum contributes nothing,
and
$$H_{-2}(x)= \frac{1}{1+x} + \cal{R}_{-2}(x) $$
Corollary 4.1 bounds the error as exponentially small
$$\cal{R}_{-2}(x) = \cal{O}\big(x^{1/2} \exp{(-D\,x^{1/3})}\big) \, , D=4.5201... $$
The proof requires some sophistication in complex analysis and some knowledge of special function theory.  I tried to solve this problem on my own, even using special function theory, and got to a place where I had to regularize some integrals.  The first term (1/x) worked with out regularization, and the rest of my terms all went to zero.  That occurrence told me that it was likely that the error was exponentially small, (that is, the asymptotic series won't contain $x^{-3/2}, x^{-2}...$) and Katsurada's paper proves it.
A: 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
$$C_i(z)=\sum_{n=1}^{\infty}c_{n,i}z^{n}/n!=zA_i'(z)=ze^z(P_i(z)-P_{i+1}(z)).$$
