$1 + 1 + 1 +\cdots = -\frac{1}{2}$

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?

Solution 1:

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 $\epsilon \rightarrow 0.$

Solution 2:

The result you obtain when calculating sums like $$S=\sum_{n=1}^\infty T_n$$ depends on how you define them. Here $T_n$ denotes anything that we may want to insert there.

The most intuitive way to define an infinite sum is by using partial sums. The idea is to introduce a sequence of sums $$S_N=\sum_{n=1}^N T_n$$ and then define the infinite sum $S$ as the following limit $$S=\lim_{N\to \infty}S_N.$$ Obviously, each partial sum $S_N$ is finite, however the problem is in this limit, that may diverge. For your example, evidently, this limit diverges and doesn't give anything useful.

To deal with this kind of sums people invented another approach called analytical continuation, that was described in the answer by Vibert. Not to repeat it I'll just say, that intuitively the idea is to consider a convergent sum instead of our divergent one. Then replace this sum by an analytical function (say Riemann zeta function). Finally, we take a limit of this analytical function in that region, where the initial sum diverges.

An example of analytical continuation is the well-known gamma function $\Gamma(n)$, that coincides with the function $(n-1)!$ when $n\in \mathbb{Z}$. However, $\Gamma(z)$ is defined for any complex $z\in\mathbb{C}$.

Solution 3:

Use following functional equation: [which is not trivial to get]

$$\pi^{-\frac{1}{2}s}\Gamma\left(\frac{1}{2}s\right)\zeta(s)=\pi^{-\frac{1}{2}(1-s)}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)$$

PS: Page 43 of the following "paper" http://www.math.ethz.ch/~gruppe5/group5/lectures/mmp/hs13/Files/Lecture%20notes%20(November%2029).pdf