Exercising divergent summations: $\lim 1-2+4-6+9-12+16-20+\ldots-\ldots$

Dear Gottfried, as you correctly observe, the sum is the Taylor expansion of $$ \frac{1}{(1+x)^3 (1-x)} $$ for $x=1$. This function has a pole at $x=1$, so the result is a genuine divergence, the standard number "infinity" (without a specification of the phase) that is inverse to zero. The fact that some seemingly divergent sums have finite values doesn't mean that all of them have finite values.

Your second method is illegitimate because it clumps the neighboring values of $n$ - the exponents in the powers of $x$ - which means that the justification isn't robust under any infinitesimal deformations of the parameters. Note that it wasn't really legitimate that you wrote $1+1+1+\dots = \zeta(0)$. In fact, $\zeta(0)-n$ for any integer $n$ - and in fact, not only integer - would be equally (un)justified. In fact, none of them gives the right result.

It's a misconception that $1+1+1+\dots = -1/2$ "always" holds. It's only true if the terms $1$ are associated with values of $n$ that go over positive integers. But if they go over non-negative integers, the result would be $+1/2$, and so on.