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I've tried to calculate few classic sums using Dirichlet regularization:

Sum[n, {n, 1, ∞}, Regularization -> "Dirichlet"]
Sum[n^2, {n, 1, ∞}, Regularization -> "Dirichlet"]

I get expected results: -1/12 and 0.

But when I start sum from the 0, I notice an interesting pattern:

$$ \sum_0^\infty n^k - \sum_1^\infty n^k = (-1)^{k+1}\frac{1}{k + 1} $$

I'm having trouble understanding this. Is this some misuse of Dirichlet regularization in Mathematica, or some interesting (or not) thing in the math itself?

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    $\begingroup$ You might want to read this Wolfram Blog article. $\endgroup$ – m_goldberg Sep 11 '14 at 14:52
  • $\begingroup$ Actually, after this article I tried this. I just can't understand what method Mathematica use; if we add zero, the Maclaurin series shouldn't change, but the answer is different. $\endgroup$ – m0nhawk Sep 11 '14 at 14:56
  • $\begingroup$ Dirichlet regularization is not stable. What that means is $\sum_{n=1}^\infty a_n \neq a_1 + \sum_{n=2}^\infty a_n$, where the sums here are Dirichlet regularized. See brilliant.org/wiki/sums-of-divergent-series/… $\endgroup$ – Chip Hurst Jan 28 '18 at 21:21
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Since Dirichlet series do not make sense when the index starts at zero, to Zeta-regularize $\sum_{n=0}^\infty f(n),$ Mathematica reasonably replaces it with $\sum_{n=1}^\infty f(n-1),$ so you are saying that $$\sum_{l=0}^{k-1} (-1)^l S_l \binom{k}{l} = (-1)^{k+1}/(k+1),$$ where $S_l$ is the regularized sum of $n^l,$ which is interesting, but is a mathematical (rather than a Mathematica) fact.

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  • $\begingroup$ Can you expand a bit more about $S_l$? And that's a binomial coefficient, right? $\endgroup$ – m0nhawk Sep 11 '14 at 16:54
  • $\begingroup$ $S_l = \sum_{n=1}^\infty n^l$ (Dirichlet-regularized). And yes, a binomial coefficient. $\endgroup$ – Igor Rivin Sep 11 '14 at 16:57
  • $\begingroup$ So, $S_l = -1/12$? It didn't calculates well: Sum[-(1/12) Binomial[k, l], {l, 1, k}] /. {k -> 3}, am I missing something? $\endgroup$ – m0nhawk Sep 11 '14 at 17:01
  • $\begingroup$ Sorry, I was typing in a hurry, look at the corrected formula (you are comparing $(n-1)^k$ with $n^k,$ so you can check this yourself...) $\endgroup$ – Igor Rivin Sep 11 '14 at 17:06
  • $\begingroup$ Now's that's will equal to $(-1)^k S_l$. Maybe you missed something else? $\endgroup$ – m0nhawk Sep 11 '14 at 17:08

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