In what sense does $\sum_{k=0}^{\infty} 2^{2k} = - {1 \over 3}$? In The Road to Reality Penrose remarks on an identity written down by Euler which is "obviously wrong" and yet correct "on some deeper level". He makes reference to the series again when discussing renormalisation.
The partial series don't look much better at $\sqrt{-4}$ than they do at 2. So the answer to making use of it can't be simply to look at $\Bbb C$. In what sense is this divergent series good?
Edit: Sorry, I mean in what sense other than as a formal power series.
Edit 2: Penrose remarks that the lower = left & right parts of the plot are "inaccessible" to the partial series. So maybe a way to rephrase the question is, how can one access them?
 A: This falls into the purview of analytic continuation. If you consider the geometric series
$$\sum_{n=0}^{\infty} x^n$$
then this converges if $|x|<1$, diverges for $|x|>1$ and it converges to $\frac{1}{1-x}$. One common way of extending this result to all of $\Bbb C$ (read: "evaluating" the sum for all $\Bbb C$) except $z=1$ is via the formal identification
$$\sum_{n=0}^{\infty} z^n \Longleftrightarrow \frac{1}{1-z}.$$
In your case, $2^{2n} = 4^n$ and so $x=4$. Substituting this into the right hand side, you get $-\frac{1}{3}$ as Penrose claims. It is not the case that the sum converges to $-\frac{1}{3}$, but more that it is a definition of sorts via the analytic continuation.
These sorts of identifications seem to give surprisingly accurate results in physics. Particularly I think the sum $\sum_{n=1}^{\infty} n$ appears in the Casimir (?) effect and renormalizing this sum via some other summation technique gives values which agree with experiment. It's not very well understood why these renormalization techniques happen to give results which agree with experiment, but with the way the theory is developed, the divergences arise almost immediately and naturally. The different summation techniques seem to mitigate the oddities.
A: The obviously wrong proof is
$$
\sum_{k=0}^{\infty}2^{2k} = \sum_{k=0}^{\infty}4^k = 1+...+4^k+... = S\\
4\sum_{k=0}^{\infty}2^{2k} = 4\sum_{k=0}^{\infty}4^k = 4+...+4^k+... = 4S
$$
subtract the two we find
$$
1 + (4-4)+...(8-8) ... = S-4S = -3S
$$
thus we get (however wrong or what not)
$$
-3S = 1\implies S = \sum_{k=0}^{\infty}2^{2k} = -\frac{1}{3}
$$
