Under this "regularization", the harmonic series can be interpreted as $s_{-1}$ and assigned a value $$s_{-1} = \lim_{k \rightarrow 1} \zeta(k) (2-2^k) = -2 \log 2$$

I was looking at regularizing divergent series to get a "reasonable" finite number. For instance, $$1+1+\cdots = - \frac12$$ can be obtained through the analytic continuation based on geometric series (or) the zeta function evaluated at the origin. Based on analytic continuation, we have that $$ 1^1 + 2^1 + \cdots = -\frac1{12} $$ $$ 1^2 + 2^2 + \cdots = 0 $$ $$ 1^3 + 2^3 + \cdots = \frac1{120} $$ $$ 1^4 + 2^4 + \cdots = 0 $$ And in general, the zeta-regularization (or) analytic continuation based on geometric series gives us $$ \begin{align} 1^k + 2^k + \cdots & = \zeta(-k) \end{align} $$ One of the first things which I hoped these infinite series would satisfy was some of the nice relations/ recurrence satisfied by their finite counterparts (the Faulhaber polynomials). For instance, $$ \begin{align} 1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( 1 + 2 + 3 + \cdots + n\right)^2 \end{align} $$ whereas $$ \zeta(-3) \neq \zeta(-1)^2 $$ I was playing around to assign a different value, (in a nice way) to the diverging series $1^3 + 2^3 + 3^3 + \cdots$ and/ or $1 + 2 + 3 + \cdots$ so that we can still have $$ \begin{align} 1^3 + 2^3 + 3^3 + \cdots = (1+2+3+\cdots)^2 \end{align} $$ and in general assigning values to the diverging series $1^k + 2^k + \cdots$ that agree with the Faulhaber's polynomial. If we define, $S(n;k) = 1^k + 2^k + \cdots n^k$ and assign a finite value to $\lim_{n \rightarrow \infty} S(n;k)$ as $s_k$, we would like $s_k = P_k(s_0)$ where $P_k(s_0)$ is a polynomial in $s_0$ of degree in $k+1$ since their finite counterparts satisfy $S(n;k) = P_k(S(n;0)) = P_k(n)$ and hence $$s_k = \lim_{n \rightarrow \infty} S(n;k) = \lim_{n \rightarrow \infty} P_k(S(n;0)) = P_k \left( \lim_{n \rightarrow \infty} S(n;0) \right) = P_k(s_0)$$ Note that $\zeta(-k) \neq P_k(\zeta(0))$ for odd $k$.

$P_k(s_0)$ is a polynomial in $s_0$ of degree $k+1$, given below. $$P_k(s_0) = \frac{s_0^{k+1}}{k+1} \sum_{q=0}^{k} \left[ \binom{k+1}{q} \left( \frac{- 1}{s_0} \right)^{q} B_q \right]$$ where $B_q$ are the Bernoulli numbers. At this juncture, note that the Bernoulli numbers also satisfy a very similar recurrence relation given by $$B_{k+1} = \frac{(-1)^{k}}{2^{k+2}-2} \sum_{q=0}^{k} \left[ \binom{k+1}{q} 2^q B_q \right]$$ given in the article here. The similarity demands that if we set $-\frac{1}{s_0} = 2$ (i.e. $s_0 = -\frac12$), we get $$ s_k = P_k \left( -\frac12\right) = \frac{\left( -\frac{1}{2} \right)^{k+1}}{k+1} \sum_{q=0}^{k} \left[ \binom{k+1}{q} 2^{q} B_q \right] $$ $$ s_k = \frac{\left( -\frac{1}{2} \right)^{k+1}}{k+1} \times \frac{2^{k+2}-2}{(-1)^{k}} B_{k+1} = - \frac{B_{k+1}}{k+1} \left( 2 - 2^{-k}\right) = \zeta(-k) \left( 2 - 2^{-k}\right) $$ Note that the regularization for $1+1+1+\cdots$ agrees with the analytical continuation i.e. $$s_0 = - \frac12 = \zeta(0)$$

Apart from what I have argued above (satisfying the Faulhaber's polynomial recurrence), is there any other reason/motivation why one would like to associate the value $(2-2^{-k}) \zeta(-k)$, instead of $\zeta(-k)$, to the diverging series $1^k + 2^k + 3^k + \cdots ?$

I initially posted this on math stackexchange where I wanted a reference for the recurrence satisfied by the Bernoulli numbers.

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    $\begingroup$ May I suggest you provide a more informative title since you're asking about one very special divergent series, not about divergent series in general? $\endgroup$ – Gerry Myerson Dec 25 '11 at 1:07
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    $\begingroup$ @Gerry: Yes. I have edited the title accordingly. Hope it is more informative now. Else feel free to edit it. $\endgroup$ – user11000 Dec 25 '11 at 1:18
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    $\begingroup$ I guess one can find a reasonable suitable definition for the "sum" of a diverging series so as to justify $1+1\dots=-1/2$ or other identity we like. However, as a next step, I would immediately abandon the notation $1+1\dots$, as it gives a sort of unnecessary air of mysticism to the whole. $\endgroup$ – Pietro Majer Dec 25 '11 at 8:40
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    $\begingroup$ Not an explanation, but the same kind of expressions appear in the Kummer congruences (en.wikipedia.org/wiki/Kummer%27s_congruence) and thus in the construction of the $p$-adic zeta function by Kubota-Leopoldt. $\endgroup$ – François Brunault Dec 25 '11 at 10:27
  • $\begingroup$ This is related, though I'm not sure if it would answer your question: mathoverflow.net/questions/3204/… $\endgroup$ – j0equ1nn Oct 19 '18 at 14:18

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