TITLE: Problem in Divergent series
QUESTION [9 upvotes]: Lets try to evaluate 
$$\frac{(-1)}{1^s}+\frac{(-1)^2}{2^s}+\frac{(-1)^3}{3^s}+\cdots$$
$$=\frac{e^{\pi i}}{1^s}+\frac{e^{2\pi i}}{2^s}+\frac{e^{3\pi i}}{3^s}+\cdots$$
$$=\frac{1}{1^s}(1+\frac{\pi i}{1!}+\frac{(\pi i)^2}{2!}+\cdots)+\frac{1}{2^s}(1+\frac{2\pi i}{1!}+\frac{(2\pi i)^2}{2!}+\cdots)+\frac{1}{3^s}(1+\frac{3\pi i}{1!}+\frac{(3\pi i)^2}{2!}+\cdots)+\cdots$$
$$=(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdots)+\frac{\pi i}{1!}(\frac{1}{1^s}+\frac{2}{2^s}+\frac{3}{3^s}+\cdots)+\frac{(\pi i)^2}{2!}(\frac{1}{1^s}+\frac{2^2}{2^s}+\frac{3^2}{3^s}+\cdots)$$
$$=\sum\limits_{k=0}^{\infty}\frac{(\pi i)^k}{k!}\zeta(s-k)$$
Now it can be shown that for $|x|<2\pi$
$$\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}\zeta(-k)=\frac{1}{1-e^x}+\frac{1}{x}-1$$
Thus
$$\sum\limits_{k=0}^{\infty}\frac{(\pi i)^k}{k!}\zeta(-k)=-\frac{1}{2}-\frac{1}{\pi}i$$
Suggesting that
$$-1+1-1+\cdots=-\frac{1}{2}-\frac{1}{\pi}i$$
Which is wrong of course, but not so wrong that is disagrees in its real part. What part of my derivation that
$$\frac{(-1)}{1^s}+\frac{(-1)^2}{2^s}+\frac{(-1)^3}{3^s}+\cdots=\sum\limits_{k=0}^\infty \frac{(\pi i)^k}{k!}\zeta(s-k)$$
is wrong and why does it give a $nearly$ correct answer?
EDIT:
Using the same method one can evaluate
$$-1+2-3+4-\cdots=-\frac{1}{4}+\frac{1}{\pi^2}i$$
Once again incorrect but correct in its real part. Coincidence? I think not!

REPLY [5 votes]: I had a very similar "surprising" result when looking at the double sum just similar to your expanding into a double sum and reordering of terms. See this link  When I encountered the problem I was astonished as you were here, and tried various heuristical explanations. However, an analytic solution was existent and so the heuristical introduction of the "infinitesimal" $\mu = 1/(-1)!$ compensated by $\zeta(1)$ like $\displaystyle w=\lim_{\delta  \to 0} {\zeta(1+\delta) \over \Gamma(\delta)}$ was somehow justified. Clearly, that use of $\mu$ should be replaced by a correct $\lim$-formula, but leave it for shortness of notation. (Similarly, sometimes the bernoulli-number $b_0$ is interpreted as that same quantity when the bernoulli-numbers are rewritten in terms of the $\zeta()$-function.) 
I just inserted in your formula that infinitesimal $\mu $ formally in your formula to work with
$$=\frac{1}{1^s}
      (\frac{\mu}{\pi i}+1+\frac{\pi i}{1!}+\frac{(\pi i)^2}{2!}+...) \\
  +\frac{1}{2^s}
      (\frac{\mu}{2 \pi i}+1+\frac{2\pi i}{1!}+\frac{(2\pi i)^2}{2!}+...) \\
  +\frac{1}{3^s}
      (\frac{\mu}{3 \pi i}+1+\frac{3\pi i}{1!}+\frac{(3\pi i)^2}{2!}+...) \\
   +...$$
getting
$$= \\ 
  \frac{\mu}{\pi i} \cdot (\frac{1}{1^{s+1}}+\frac 1{2^{s+1}}+\frac 1{3^{s+1}}+...) \\
  +1                \cdot (\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+...) \\
  +\frac{\pi i}{1!} \cdot (\frac{1}{1^s}+\frac{2}{2^s}+\frac{3}{3^s}+...) \\
  +\frac{(\pi i)^2}{2!}(\frac{1}{1^s}+\frac{2^2}{2^s}+\frac{3^2}{3^s}+...) \\
  + $$
where -if you let $s \to 0$ the first summand expands now to
 $$ {\mu \over \pi i} \cdot \zeta(1) = \lim_{\delta \to 0} {\zeta(1+\delta) \over \Gamma(\delta) } \frac 1{\pi i} = \frac 1{\pi i}
$$
and with that correction term you come back to the correct value in your final formula.
This heuristic fumbling is surely not formally correct but may give a hint how to repair your formula: I think usually we'll arrive at some additional integral-expression instead of the infinitesimal (likely by the Euler-MacLaurin-formula), or you could try to re-develop your formula using the Bernoulli-polynomials where the hack which I've indicated here with the $\mu$ is captured in the bernoulli-number $b_0$.       
Sorry I can't do the complete process in the correct way, but perhaps this is instructive/helpful anyway.