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 $$ \lim_{n\to\infty}\left(s_2(n)-n\frac{\pi^2}6+\log n\right)=\frac{\pi^2}6-1-\gamma \qquad \text{introduced by me: } =\zeta(2)-1 - \gamma $$

Regularizing the sum when the analytically continued function has a pole at the value of interest for the parameter, does not seem to give a unique value. There are several options to regularize:

1. You could introduce a variable $x$ to make it a power-series s.t. the series converges for $|x|<1$ i.e. $$\sum_{k=1}^\infty x^k \sum_{j=1}^k \frac{1}{j^2} = \sum_{j=1}^\infty \frac{1}{j^2} \sum_{k=j}^\infty x^k = \sum_{j=1}^\infty \frac{1}{j^2} \frac{x^j}{1-x} = \frac{{\rm Li_2}(x)}{1-x} \, .$$ Finally expand the polylogarithm about $x=1$ using the polylogarithm identity https://dlmf.nist.gov/25.12 mentioned in the comments, which gives $$\frac{{\rm Li_2}(x)}{1-x} = \frac{-{\rm Li_2}(1-x) + \zeta(2) - \log(x)\log(1-x)}{1-x} \\ = \frac{\zeta(2)}{1-x} + \log(1-x) - 1 + {\cal O}\left( (1-x)\log(1-x) \right)$$ for $x\rightarrow 1$. Hence, when the bare singularities are discarded you arrive at $-1$.

2. Cut-Off regularization introduces an upper limit $m$ into the sum and the result is then expanded in terms of the cut-off. This gives $$\sum_{k=1}^m H^{(2)}_k = (m+1) H_m^{(2)} - H_m = (m+1)\zeta(2) - \log (m) - \gamma - 1 + {\cal O}(1/m) \, .$$ The divergent parts in $m$ can be viewed as singularities at $\infty$ and are of the same type as in the first method for $x\rightarrow 1^{-}$. The constant term differs and reads $\zeta(2) - \gamma - 1$.

3. Zeta functional regularization is another commonly applied procedure to treat infinities. The generalized zeta-function which converges for $\Re(s)>1$ is sought to be continued to $s=0$. Since $s=1$ has convergence issues, the analytically continued version will turn out to have a pole at $s=1$. Therefore we first continue to $\Re(s)>0$ by simple manipulation \begin{align} \sum_{k=1}^\infty \frac{H_k^{(2)}}{k^s} &= \sum_{k=1}^\infty k^{-s} \sum_{j=1}^k j^{-2} = \sum_{k=1}^\infty k^{-s} \left( \sum_{j=1}^\infty j^{-2} - \sum_{j=k+1}^\infty j^{-2} \right) \\ &= \zeta(2) \zeta(s) + \zeta(s+2) - \sum_{k=1}^\infty k^{-s} \sum_{j=k}^\infty j^{-2} \\ &= \zeta(2) \zeta(s) + \zeta(s+2) - \sum_{j=1}^\infty j^{-2} \sum_{k=1}^j k^{-s} \, . \end{align} The Zeta-function of the first term has the aforementioned pole at $s=1$ and can be easily continued to $s=0$ and the rightmost sum is now convergent for $\Re(s)>0$ which implies there to be another pole at $s=0$. To shift the convergence further to the left in the complex plane we now introduce $$k^{-s} = \frac{1}{\Gamma(s)} \int_0^\infty x^{s-1} e^{-kx} \, {\rm d}x$$ into the last term, which in turn becomes (after interchanging summation and integration) \begin{align} \sum_{j=1}^\infty j^{-2} \sum_{k=1}^j k^{-s} &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \sum_{j=1}^\infty j^{-2} \sum_{k=1}^j e^{-kx} \\ &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \sum_{j=1}^\infty j^{-2} \, \frac{1-e^{-jx}}{e^x-1} \\ &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \, \frac{\zeta(2)-{\rm Li_2}\left(e^{-x}\right)}{e^x-1} \\ &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \, \frac{{\rm Li_2}\left(1-e^{-x}\right)-x \log\left( 1-e^{-x} \right)}{e^x-1} \, , \end{align} where in the last step the aforementioned polylogarithm identity was used. The integral converges like the sum for $\Re(s)>0$ and each term of the integral converges separately. The restriction follows from the behaviour at $x=0$. Because of the $\frac{1}{\Gamma(s)}$ it suffices to extract the singularities at $s=0$, since any finite contribution will vanish in the limit $s\rightarrow 0$. To do this, we subtract and add the leading behaviour of the nominator for $x\rightarrow 0$ \begin{align} {\rm Li_2}\left(1-e^{-x}\right) &= x + {\cal O}(x^2) \\ \log\left(1-e^{-x}\right) &= \log(x) + {\cal O}(x) \, . \end{align} After the leading terms have been subtracted, the resulting integral converges for $\Re(s)>-1$, i.e. it is finite at $s=0$. Hence it suffices to calculate $$ \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \, \frac{x-x \log\left( x \right)}{e^x-1} \\ = \frac{s}{\Gamma(s+1)} \int_0^\infty \frac{x^s}{e^x-1} \, {\rm d}x - \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^s}{e^x-1} \, \log(x) \, {\rm d}x \\ = s \, \zeta(s+1) - \frac{1}{\Gamma(s)} \frac{{\rm d}}{{\rm d}s} \Gamma(s+1)\zeta(s+1) \\ = 1 + \frac{1}{s} + \gamma + {\cal O}(s) $$ where the integral representation of the Zeta function was used from the second to the third line. Evidently there is no logarithmic singularity like in the other two cases, but they all share the leading pole singularity. The constant term in this case after using the well known result $\zeta(0)=-\frac{1}{2}$ finally reads $\frac{\zeta(2)}{2} - \gamma - 1$.

What's the moral of the story? When the analytically continued version of the to be regularized sum contains a pole at the value in question, the result can be more or less anything. Only when the analytically continued value is finite, it is also unique. You can also see the arbitrariness for instance when you substitute $m \rightarrow m+c$ in the second case, or $x \rightarrow \frac{x + c (1-x)}{1+c(1-x)}$ in the first case, for any constant $c$.

Similair as above but more in line with the rest of the picure you posted

The eulerian numbers(s,n): $\Delta^{s+1}n^s$ cut-off at 1. (so you ignore 0 and bellow when delta'ing).

Upper limit when s is positive integer s btw.

$\sum_{n=1}^{\infty} \frac{ Eulerian.numbers(s,n) c^n}{(1-c)^{s+1}}=\sum_{n=1}^{\infty}c^n n^s$

Now at the negative side:

Eulerian numbers(-s,n): $.^{s-1}\sum_{m=1}^{n}m^{-s}$ (the s'th -1 'indefinite' sum, I aways struggled with notation for this cause it's just similair to a multiple integral).

In your case with s=-2. Eulerian numbers(-2,n): $\sum_{n=1}^{k}n^{-2}$.

$\sum_{n=1}^{\infty} \sum_{n=1}^{k}n^{-2} c^n(1-c)=\sum_{n=1}^{\infty}c^n n^{-2}$

You can "see it" by (1-c) which works like a delta operator. $$\sum_{n=1}^{\infty} \sum_{n=1}^{k}n^{-2} c^n=\frac{\sum_{n=1}^{\infty}c^n n^{-2}}{1-c}$$

From here on figure the limit as c goes to 1, by expending the polynomal. What the answer previous stated before was that you could disregard the divergent parts, which is I think incorrect, e.g. zeta(1) which is also here the case I think.

An intuitive example:

$$\sum_{n=1}^{\infty} \frac{c^n}{n^2 (1-c)}$$ Let c goes to 1+h with h being small. Only looking at the $h^0$ order gives $$\sum_{n=1}^{\infty} \frac{(1+h)^n}{n^2 (-h)}= \sum_{n=1}^{\infty} \frac{-1}{n}=-\zeta(1)$$

Now like you normally would (or see directly from it series it's going to contain a full zeta(1)).

$$\sum_{k=1}^{\infty} \sum_{n=1}^{k}n^{-2}=\sum_{k=1}^{\infty}\pi^2/6-\frac{1}{k}+\frac{1}{2k^2}-\frac{1}{6k^3}+\frac{1}{30k^5}-...$$

$$\sum_{n=1}^kn^{-2}=\bigg(\pi^2/6-\sum_{j=-1}^{\infty}\zeta(-j)(k^{-2-j}) \frac{(-2)!}{(-2-j)!(j)!}\bigg)$$

$$\sum_{k=1}^{\infty} \sum_{n=1}^{k}n^{-2}=-\pi^2/12-\sum_{j=-1}^{\infty}\zeta(-j)\zeta(2+j) \frac{(-2)!}{(-2-j)!(j)!}$$ And as you see $\zeta(1)$ is part of the sum at j=-1 twice and it will only gets reduced once by (-1)!. There for the regularized value got a pole, e.g. contains zeta(1).

Or better writen:

$$\sum_{k=1}^{S\infty} \sum_{n=1}^{k}n^{-2}=-\zeta(1)-\sum_{j=1}^{\infty}\zeta(-j)\zeta(2+j) (-1)^{j+1} (j+1) $$

And even though the other sum also diverges it seems unlikely it contains zeta(1) but yeah it needs a more convincing argument/needs to be regularized first aswell. But it seems to be regulaziable.

Ps: doesn't it alternate in the picture? And thuse is trivial. It's used for c isn't 1 to very easily calculate the zeta function. I'd say it's the basics. Cause if it doesn't alternate you also get for sum 1/n just zeta(1) in the entry before.