i tried to add the formula $\sum_{n=1}^{100}\frac{1}{x^{n}}$ as to aproximate the value as x = 2 which equals to 1 as most of you might know i graphed it at desmos and this happened
as you can see the value of x=1 gives y = 0 to farther explain$\sum_{n=1}^{100}\frac{1}{-1^{n}}$ which equals to :
$\frac{1}{\left(-1\right)^{1}}+\frac{1}{\left(-1\right)^{2}}+\frac{1}{\left(-1\right)^{3}}+\frac{1}{\left(-1\right)^{4}}$...
=-1+1-1+1-1+1...
which equals zero(on the graph) even though it diverges in reality alternating bettween 0 and -1
why is that?
We know that $\sum_{n=1}^{100} \frac{1}{(-1)^n}$ (note: be careful, you forgot the parentheses) is equal to
$$\frac{1}{(-1)^1}+\frac{1}{(-1)^2}+\frac{1}{(-1)^3}+\cdots+\frac{1}{(-1)^{100}} = -1+1-1+1-1+1...-1+1.$$
Since $100$ is an even number, we get exactly 50 $-1$s and 50 $1$s, which cancel out to become $0$. If we tried an odd number, say $99$, we would have $-1$ instead of $0$. Of course, at infinity, the sum diverges between $0$ and $-1$, just as you said.