Is this true?

[dreamy geyser]

Iam extremely sorry

This is the actual one

[paper breach]

basically yes

[dreamy geyser]

thanks but

prove it

[paper breach]

theres a bunch online so yeah

[dreamy geyser]

got it

[timid geyser]

It’s too difficult to write the prove there on discord

A lot of fractions

[dreamy geyser]

hmmmmmmmmmmmm

:|

OH

shi

i think i found something

$\sum_{r=1}^{\infty}\left(\frac{1}{n^{r}}\right)=\sum_{n=1}^{\infty}\left(n^{-r}\right)=\frac{1}{n-1}=\left(n-1\right)^{-1}=-\left(1-n\right)^{-1}=-\left(1+\left(-1\right)\left(-n\right)+\frac{\left(-1\right)\left(-1-1\right)}{2!}\left(-n\right)^{2}+\frac{\left(-1\right)\left(-1-1\right)\left(-1-2\right)}{3!}\left(-n\right)^{3}...\right)=-\left(1+n+n^{2}+n^{3}...\right)$

[hard smelt]

kev

[dreamy geyser]

now all are negative...

[timid geyser]

The power series (1+x)^-1 is only available for x<1

[dreamy geyser]

It's (1-x)^-1

[timid geyser]

Same thing 🥲