# Another Taylor series question

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• Apr 3rd 2008, 09:02 PM
emttim84
Another Taylor series question
I don't know, maybe it's late and I just need to knock off, but I can't quite nail the power series for the function $\displaystyle \frac{4}{x-1}$ with the center at zero. $\displaystyle \frac{1}{x-1}$ is $\displaystyle (-1)^nx^n$, however, so could I simply put the 4 out in front as a coefficient and use that? $\displaystyle 4(-1)^nx^n$ perhaps?
• Apr 4th 2008, 02:56 AM
Mathstud28
Here
Yes the power series would be $\displaystyle \sum_{n=0}^{\infty}4(-1)^{n}x^{n}\$...since all you do is say that $\displaystyle \frac{4}{1-x}$=$\displaystyle 4\frac{1}{1-x}$=$\displaystyle 4\sum_{n=0}^{\infty}(-1)^{n}x^{n}$ and distribute the 4 to get $\displaystyle \sum_{n=0}^{\infty}4(-1)^{n}x^{n}\$
• Apr 4th 2008, 08:16 AM
colby2152
Just multiply it by 4.

Think... $\displaystyle \sum x = 1 + 2 + 3 + 4 + 5 \ldots$

$\displaystyle 4 \sum x \Rightarrow \sum 4x = 4(1) + 4(2) + 4(3) + 4(4) + 4(5) + \ldots \Rightarrow 4(1 + 2 + 3 + 4 + 5) + \ldots$
• Apr 4th 2008, 09:11 AM
emttim84
Ok, so that is all I do...thanks guys!