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Mathematics 18 Online
OpenStudy (anonymous):

Determine a function which is represented by that summation given below

OpenStudy (anonymous):

\[\sum _{n=1}^{\infty } \text{nx}^n\]

OpenStudy (anonymous):

ok i guess we will worry about the radius of convergence second. this looks almost like a derivative, so we will treat it as one.

OpenStudy (anonymous):

\[\Sigma_1^\infty x^n=\frac{1}{1-x}\] for -1<x<1 yes?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

taking derivatives we get \[\Sigma_1^\infty nx^{n-1}=\frac{1}{(1-x)^2}\]

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

multiply both sides by x to get \[\Sigma_1^\infty nx^n=\frac{x}{(1-x)^2}\]

OpenStudy (anonymous):

i have been very sloppy here, especially from where we started. i think we have to be careful with starting at n = 0 or n = 1, but i will let you worry about this. i will also let you worry about the validity of differentiating a power series term by term and the radius of convergence. but the general idea is there and i think the answer is correct modulo those details.

OpenStudy (anonymous):

Okay,thanks

OpenStudy (anonymous):

i think in fact the very first line i wrote was incorrect. i think \[\Sigma_0^\infty x^n = \frac{1}{1-x}\] no what i wrote.

OpenStudy (anonymous):

but of course when n = 0 in your power series you get 0 anyway, so you might as well start at 1!

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