False proof that $1-1+1-1+1\cdots= \frac12$

Question

The expression $1+\dfrac{1}{x}+\dfrac{1}{x^2}+\cdots$ is equal to $\dfrac{x}{x-1}$ when $|x|>1$ by using the formula for geometric series. But the limit as $x\rightarrow-1$ of the expression is equal to $\frac12$ (this can be clearly seen on a graph). However the limit as $x\rightarrow-1$ of the expression is also equal to $1-1+1-1+1+\cdots$, so therefore $1-1+1-1+1\cdots=\dfrac{1}{2}$?

I feel as though I messed up my limits somewhere, can someone help me please?

Answer 1

Let me explain, why your approach does not work: We have for $|x|>1$ $\sum_{n=0}^\infty1/x^n=\frac{x}{x-1}$, it is absolutely correct that the limit from the left: $\lim_{x\to -1^{-}}\sum_{n=0}^\infty1/x^n=1/2$, but this does not mean that $1-1+1-1+\cdots$ converges to $1/2$ because you can not naively interchange sum and limit.

Answer 2

$$ \int_{0}^{1}xdx = \frac{1}{2}$$ $$ \int_{0}^{1}xdx = \int_{0}^{1}e^{\ln(x)}dx $$ $$ \int_{0}^{1}\sum_{0}^{\infty} \frac{\ln^{n}(x)}{n!}dx = \sum_{0}^{\infty}\frac{1}{n!} \int_{0}^{1} \ln^{n}(x)dx $$ $$\sum_{0}^{\infty}\frac{(-1)^{n}n!}{n!} = \sum_{0}^{\infty}(-1)^{n}$$ This does not mean that this series is equal $\frac{1}{2}$. This mean that the result of this development is a divergent series. So, for some pre-defined scopes, some divergent series could have some values. If and only if you specify the value and scope in logic terms.