Can someone demo how to find the sum to infinity of a series like this:
[eqn]1+ \frac{1} {x} + \frac{1} {x^2} + \frac{1} {x^3}+ \frac{1} {x^4}+... [/eqn]
?
>pic unrelated
wiki geometric series, and its derivation, and there substitute
x for 1/x
>>7987631
Fucking brits and their authoritarian behaviour
>>7987651
>msu.edu
>Michigan State University
You do see the dollar signs on the left of the poster, right? Last I checked Britain used the pound.
>>7987659
Depends on the x
Okay, so I am working through the question in the picture, but my answer doesn't match up with the model one and I dont see why.
My working:
[eqn]x+2<\frac {x}{x-1}
\\(x-1)(x+2) < x
\\x^2 - 2 < 0
\\ \sqrt{2} < x < \sqrt{2}[/eqn]
but the question specifies x > 0 therefore
[eqn]x<\sqrt{2}
[/eqn]
Unfortunately this is not correct, and I don't know why. I can see the model answer multiplied by (1-x) squared, but I don't know why.
How did he know to do that?
>>7987668
whoops, that final answer should obviously read
[eqn]0<x< \sqrt{2} [/eqn]
and i meant the model answer multiplied by (x-1) squared, not (1-x)
my bad
>>7987657
>You do see the dollar signs on the left of the poster, right?
I didn't.
Even worse that posters like this exist in burgerland too.
hey OP do you go to MSU? I do and live in cedar village lets have a 4chan meet
>>7987651
LOL it's from Michigan State and it's about riots after sports games where drunk idiots flip cars and burn furniture.
>>7987631
>Michigan State
Where are you from?
>>7987668
Because (x-1)^2>0 and we are dealing with an inequality. If we multiply by (x-1) we don't know if it will flip it or not because we don't know the sign of x.
