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.
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 limz→1 of the sum
0+1z1+2z2+3z3+…
diverges, because of the pole in
∑∞k=0kzk=zddz∑∞k=0zk=zddz11−z=z(z−1)2,z∈(0,1)
We are instead going to consider the sum of smooth deviations of the above, using the local mean
⟨f(k)⟩:=∫k+1kf(k′)dk′.
for which ⟨kzk⟩=zddz⟨zk⟩=zddz⟨eklog(z)⟩=zddzzk′log(z)|k+1k.
Because of canceling upper and lower bounds, the sum ∑nk=0⟨kzk⟩ is z0log(z)2 plus terms suppressed by zn. Finally, using the expansion
1(log(1+r)/r)2=11−r+(1−11!2!3!)r2+O(r3)=1+r+11!2!3!r2+O(r3),
we find
∑∞k=0(kzk−⟨kzk⟩)=z(z−1)2−1log(z)2=−112+O((z−1)1).
The picture shows the two functions z(z−1)2 and 1log(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
−11!2!3!=−112=−0.08˙3.
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