Grandi's series,
$$1+1-1+1-1+1-1+…$$
can be expressed as the below:
$$\sum_{n=0}^\infty(-1)^n$$
Two valid sums that make sense to me are $1$, and $0$, depending on how you approach the series. $(1+1)-(1+1)-(1+1)-…=0$, and $1+(1-1)+(1-1)+(1-1)+…=1$.
There is consensus, however, that the actual sum is $\frac{1}{2}$. Why? I understand the approach of finding partial means of the series, and they do indeed tend to $\frac{1}{2}$, but it seems unintuitive to assert that the sum is neither $1$ or $0$.
A more convincing method I found was assuming the series is $S$, then shifting it such that $S-1 = S$, then through algebra finding $S = \frac{1}{2}$, but again, it seems more intuitive answer is either $0$ or $1$. I say this strictly because adding and subtracting integers should equal an integer, never a fraction.
Is this a characteristic of infinite series, which is not specific to Grandi's series?
Best Answer
Consider power series $$ S(x) = \sum_{n=0}^{\infty} (-1)^n x^n, \qquad x\in [0;1). $$ It is geometric series: $$ \sum_{n=0}^{\infty} (-x)^n = \frac{1}{1-(-x)} = \frac{1}{1+x}. $$
So, $$ S(x)=\frac{1}{1+x}, \qquad x\in[0,1). $$
$S(x)$ is continuous and bounded on $[0;1)$. So, we can find limit: $$ S = \lim_{x\to 1} S(x) = \frac{1}{2}. $$
See Abel summation for better understanding.