I have already asked a similar question about Ramanujan's summation in general but received no good answers. Now I am interested in this exact series:
$$\sum _{n\ge1}^\Re (24 n + 12 n^2)$$
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$\begingroup$ what is $\mathfrak R$? $\endgroup$– glSCommented Aug 29, 2015 at 6:41
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1$\begingroup$ @glance, it's a notation indicating that the sum should be interpreted as a Ramanujan sum. $\endgroup$– J. M.'s missing motivation ♦Commented Aug 29, 2015 at 6:47
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$\begingroup$ Anyway, did you already try using the formulae in your other post on $x^n$? $\endgroup$– J. M.'s missing motivation ♦Commented Aug 29, 2015 at 6:50
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1$\begingroup$ @Guesswhoitis. Sumthing like that ;-} $\endgroup$– ciaoCommented Aug 29, 2015 at 6:50
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$\begingroup$ @Guess who it does not resolve symbolically. $\endgroup$– AnixxCommented Aug 29, 2015 at 7:44
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2 Answers
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From solution provided by kirma to your previous question
Sum[24 n + 12 n^2, {n, 1, Infinity}, Regularization -> "Dirichlet"]
(* -2 *)
From solution provided by xzczd to your previous question
ramanujanSum[f_] :=
Block[{x, n},
FullSimplify[-Sum[
BernoulliB[n, 1]/n SeriesCoefficient[
f[x], {x, 0, n - 1}], {n, \[Infinity]}], n >= 1]]
ramanujanSum[24 # + 12 #^2 &]
(* -2 *)
From Wikipedia $$1+2+3+\cdots +=-\frac{1}{12}(R)$$
Extending to positive even power, this give:
$$1+2^{2k}+3^{3k}+\cdots +=0(R)$$
24*(-1/12) + 12*(0)
(* -2 *)
answered Aug 29, 2015 at 15:10
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$\begingroup$ @Anixx - since it gives the same result as other approaches shown in comments and other answers, I assume that you have a problem in general with Ramanujan summations. $\endgroup$ Commented Jun 2, 2016 at 17:12
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Using the Zeta function you can calculate arbitrary expression like in your example.
g = 24 Zeta [-1 ] + 12 Zeta [-2]
(* -2 *)
answered Aug 30, 2015 at 12:38