Regularized sum of $1+(1+\frac1{2^2}) + (1+\frac1{2^2}+\frac1{3^2})+\cdots$, or $\sum_{k=1}^\infty \sum_{j=1}^k\frac1{j^2}$? For my exercises on divergent summation I try to find a regularization of
$$   S^{^\star}_2 \underset{\mathcal R}= \sum_{k=1}^\infty \sum_{j=1}^k \frac1{j^2} \tag 1
$$
It is not difficult to derive the formula for the partial sums introducing the harmonic numbers of order $1$ and $2$
$$H_1(N)=\sum_{k=1}^N \frac1{k} \\
  H_2(N)=\sum_{k=1}^N \frac1{k^2} \tag {2.1}$$
and writing
$$ S_2(N) = (N+1) \cdot H_2(N) - H_1(N) \tag {2.2}$$
When going to the limit $N \to \infty$ and the serial summation according to eq 2.1 cannot be done, we can -using the $\psi$ (or "digamma", in Pari/GP  psi(1+N))-function-   approximate  $$H_1(N) \sim \gamma + \psi(1+N)  \tag {3.1}$$  and for $H_2(N)$ the derivative of the $\psi()$:
$$ H_2(N) \sim \zeta(2)-\psi^{(1)}(1+N) \tag {3.2}
$$
I don't see how I can proceed now. I'd assume for the regularization in the case for $N$ in the limit I can use $\gamma$ for the $H_1^{^\star}$.

Q:But how can I the do the $H_2$-part for the limit formula for eq (3.2) resp. (2.2)?


A bit background: In a treatize on the triangle of Eulerian numbers (pg.8) I hand-waving-generalize the triangle towards negative row numbers, and for the row-sum of the $-2$'nd row I get those formal sums.


update I just found my 2013-question about the same subject, and there was an answer but which I could not then believe it is a final answer, and still I'm unsure whether (and how) I can proceed to a "fixed statement" see earlierquestion The user @did gave the answer
$$
\lim_{n\to\infty}\left(s_2(n)-n\frac{\pi^2}6+\log n\right)=\frac{\pi^2}6-1-\gamma \qquad \text{introduced by me: } =\zeta(2)-1 - \gamma
$$

Your question doesn't make much sense.
For $\Re(s) > 3$
$$\sum_{n\ge 1} ((n+1)^{3-s}-n^{3-s}) \sum_{m=1}^n m^{-s} = -\zeta(2s-3)$$
Then $$\lim_{s\to 2} -\zeta(2s-3)= \infty$$
but $$\lim_{t\to 0^+}-\Re(\zeta(2(2+it+t^3)-3)=-\gamma$$
