Are all infinite sums not divergent? In quantum field theory
No, many infinite sums converge. An infinite sum (or "series") $a_0 + a_1 + a_2 + \dots$ is defined to converge to a value $S$ if the limit $$ S = \lim_{n\rightarrow \infty} \sum_{i=1}^n a_i$$ exists. For example, if the value of $a_i$ falls off exponentially quickly or as a power law faster than $1/i$, then the series converges. The various convergence tests you learned in calculus can give you more precise criteria for convergence.
The infinite series $1 + 1 + 1 + \dots$ is not convergent - the above limit does not exist. However, it is regularizable - that is, you can play some tricks on it that "beat it into shape" well enough that you can assign some finite number to it. But this finite number is not actually the sum, which does not exist. Being regularizable is a much weaker criterion than being convergent.
So whenever you come across a divergent series in QFT and replace it with its regularized value, it's very important that you take into account that the two quantities aren't actually equal. Despite its being very important, there are approximately zero physicists who actually do it.
Edit: the OP asked a very good question in the comments that I though was worth addressing in my main answer: whether imposing different regulators on the same divergent series always yields the same result. If anyone has any thoughts, I've posed that question at Can different choices of regulator assign different values to the same divergent series?. Also, I once asked a related question at https://physics.stackexchange.com/questions/254051/how-can-dimensional-regularization-analytically-continue-from-a-discrete-set for which I never got a satisfactory answer.
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 $1+1+1+\dots=-1/2$.