Read more carefully the first part of Terry Tao's post. He replaces the regular partial sums with a smoothed sum

$$\sum_{n=1}^N n^s \to \sum_{n=1}^\infty \eta(n/N) n^s$$

where $\eta$ is a *cutoff function*. The result he finds for the smoothed sum of $1+1+1+\dots$ (case $s=0$) is

$$\sum_{n=1}^\infty \eta(n/N) = - \frac 1 2 + C_{\eta,0} N + O(1/N)$$

where

$$C_{\eta,0} =\int_0^\infty \eta(x) dx$$

On the right side there is an *asymptotic expansion* of the smoothed sum. As you can see, there is indeed a constant term $-1/2$, but the following term is *divergent* in the limit $N \to \infty$. So it is misleading to state that

$$1+1+1+\dots = -1/2$$

because $-1/2$ is just the constant term of an asymptotic expansion which is divergent in the limit $N\to \infty$.

What is usually done in QFT (see Luboš Motl's answer here for example) is to cancel the leading divergence by means of a local counterterm. Basically, a "trick" is used to ged rid of the divergence and be able to write $1+1+1+\dots=-1/2$.

This post imported from StackExchange Mathematics at 2016-07-10 19:38 (UTC), posted by SE-user valerio92