If D is critical dimension of Bosonic strings, a particular derivation goes like the following, where we arrive finally at D−22∞∑n=1n+1=0. Now mathematically this is clearly a divergent series, but using zeta function regularization here we are taking ∞∑n=1n=ζ(−1)=−112. And obtain D=26 where ζ is the analytic continuation of the zeta function we know. But it makes no sense in putting s=−1 in the formulae ζ(s)=∞∑n=11ns. As the above is only valid for Re(s)>1. So what is going on in here? Can anyone give me a reasonable explanation about obtaining −1/12?
Answer
I know some derivations in which one can track the emergence of the concrete value, without having to buy that the second order contribution in the Euler-MacLaurin formula (see other answer) is −12! times the second Bernoulli number B2.
The limit lim of the sum
0+1\,z^1+2\,z^2+3\,z^3+\dots
diverges, because of the pole in
\sum_{k=0}^\infty k\,z^k=z\frac{{\mathrm d}}{{\mathrm d}z}\sum_{k=0}^\infty z^k=z\frac{{\mathrm d}}{{\mathrm d}z}\frac{1}{1-z}=\frac{z}{(z-1)^2}, \hspace{1cm} z\in(0,1)
We are instead going to consider the sum of smooth deviations of the above, using the local mean
\langle f(k)\rangle:=\int_{k}^{k+1}f(k')\,{\mathrm d}k'.
for which \langle k\,z^k\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\langle z^k\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\langle {\mathrm e}^{k \log(z)}\rangle=z\frac{{\mathrm d}}{{\mathrm d}z}\frac{z^{k'}}{\log(z)}\left|_{k}^{k+1}\right..
Because of canceling upper and lower bounds, the sum \sum_{k=0}^n\langle k\,z^k\rangle is \frac{z^0}{\log(z)^2} plus terms suppressed by z^n. Finally, using the expansion
\dfrac{1}{\left(\log(1+r)\,/\,r\right)^2}=\dfrac{1}{1-r+\left(1-\frac{1}{1!\,2!\,3!}\right)r^2+{\mathrm{O}}(r^3)}=1+r+\dfrac{1}{1!\,2!\,3!}r^2+{\mathrm{O}}(r^3),
we find
\sum_{k=0}^\infty \left(k\,z^k-\langle k\,z^k\rangle\right)=\dfrac{z}{(z-1)^2}-\dfrac{1}{\log(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right).
The picture shows the two functions \dfrac{z}{(z-1)^2} and \dfrac{1}{\log(z)^2}, as well as their difference (blue, red, yellow). While the functions themselves clearly have a pole at z=1, their difference converges against
-\frac{1}{1!\,2!\,3!}=-\frac{1}{12}=-0.08{\dot 3}.
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