Is it logical?

The formula for the Riemann zeta function $\zeta(s) \; = \; \sum_{n-1}^\infty \frac{1}{n^s}$ is only convergent for $\mathrm{Re}\,s > 1$ . We need to apply analytic continuation to show that $\zeta(s)$ has a meaning for other values of $s$ , and eventually deduce that $\zeta(-1) = -\tfrac{1}{12}$ . This does not make the logic of substituting $s=-1$ into the first formula valid, since the formula $\sum_{n=1}^\infty n$ is divergent. A very different expression for $\zeta(s)$ has to be used to evaluate $\zeta(-1)$ .

Here is an analogy. The infinite GP formula $\sum_{n=0}^\infty x^n ;\ = \; \frac{1}{1-x}$ is valid for $-1 < x < 1$ , but the formula on the right makes sense for (for instance) $x=2$ . This does not mean that $\sum_{n=0}^\infty 2^n \; = \; -1$ The series $\sum_{n=1}^\infty2^n$ is divergent.

It is true that it is possible to make correct deductions in some parts of particle physics by using "results" like $\sum_{n=1}^\infty n = -\tfrac{1}{12}$ . This is simply because that "result" is being used a a short-hand for a proper application of Riemann zeta regularization. The incorrectness of the "result" still stands, and the supposed logic that uses it without comment is invalid.