Consider the zeta function $\zeta(s)= \sum \limits_{n=1}^{\infty} \frac{1}{n^s}$.
It is established that $ \zeta(-1) = -\frac{1}{12}$.
Reference (Equation 90)
Then we have $ \zeta(-1) = \sum \limits_{n=1}^{\infty} \frac{1}{n^{-1}}= 1+2+3+4 + ... = -\frac{1}{12}$.
But of course this series is divergent. So what is the problem here?
Power Series are only valid inside a radius of convergence. For example, the geometric series
$$1 + z^2 + z^4 + z^6 + \cdots \equiv \frac{1}{1-z^2}$$
for all $|z| < 1$. The identity does not hold for $|z| > 1$, and the identity may or may not hold when $|z|=1$. You need to read about analytic continuation.