Is this true? $\sum_{n=0}^\infty\left(-1\right)^n=\sum_{n=0}^\infty\left(-1\right)+\sum_{n=0}^\infty\left(1\right)$

Question

Is this statement about Grandi's series true? $$ \sum_{n=0}^{\infty }\left ( -1 \right )^{n}= \sum_{n=0}^{\infty }\left ( -1 \right ) + \sum_{n=0}^{\infty }\left ( 1 \right ) $$

I was playing with infinite sums, and I noticed that since $\sum_{n=0}^{\infty }\left ( -1 \right )^{n}$ is just 1 - 1 + 1 - 1 + 1..., and can be separated into two series. I know that Grandi's series is divergent, but can it still be a sum of two other series?

Answer 1

There is no infinite grouping of terms allowed if a series is divergent. So you can not write a divergent series as a sum of two other divergent series.

Answer 2

No, this is not true in any meaningful sense.

The series $\sum_{n=0}^{\infty }\left ( -1 \right )^{n}$ has (Cesaro, Dirichlet) regularized value of $-1/2$, the series $\sum_{n=0}^{\infty }\left ( -1 \right )$ has (Dirichlet) regularized value of $-1/2$ and $\sum_{n=0}^{\infty }1$ has regularized value $1/2$. So, the right hand side has regularized value of $0$ (regularization is a linear operator), and the left-hand side has the regularized value of $-1/2$, so they cannot be equal.