sequences and series - $1 + 1 + 1 +cdots = -frac{1}{2}$

The formal series

$$\sum_{n=1}^\infty 1 = 1+1+1+\dots=-\frac{1}{2}$$

comes from the analytical continuation of the Riemann zeta function $\zeta (s)$ at $s=0$ and it is used in String Theory. I am aware of formal proofs by Prof. Terry Tao and Wikipedia, but I did not fully understand them. Could someone provide an intuitive proof or comment on why this should be true?

Answer

Let me walk you through the Riemann zeta computation. Call $S$ your original sum. Let's regulate the sum as follows:

$$S_s \equiv \sum_{n \geq 1} \frac{1}{n^s}.$$
Fix $n \geq 1.$ Then $n^{-s} \rightarrow 1$ as $s \rightarrow 0,$ so if we can assign a meaning to $S_s$ as $s \rightarrow 0$, we can interpret $S$ as this limit.

Now, for $s > 1$ the above sum exists and it equals the Riemann zeta function, $\zeta(s).$ $\zeta$ has a pole at $s=1$, which is just the statement that the (non-regulated) sum $\sum 1/n$ diverges. But we can analytically continue $\zeta$ if we take care to avoid this pole. Then we can Taylor expand around $s=0$

$$\zeta(s) = -\frac{1}{2} - \frac{1}{2} \ln(2\pi) s + \ldots$$
which implies that

$$S = \lim_{s \rightarrow 0} S_s = -\frac{1}{2}.$$
(The equality sign is to be understood in the regulated sense.)

There are many other ways to regulate the sum. You can e.g. suppress the tail as $\sim \exp(-\epsilon n)$, but then you need to add a counterterm to absorb a pole as $\epsilon \rightarrow 0.$