An Interesting Infinite Geometric Series

**icemanfan** · January 13th 2012, 09:08 PM

Express the infinite series 1 + (1/2)x + (1/3)x^2 + (1/4)x^3 + (1/5)x^4 + ... as an elementary function of x.

**CaptainBlack** · January 13th 2012, 09:42 PM

Originally Posted by icemanfan

Express the infinite series 1 + (1/2)x + (1/3)x^2 + (1/4)x^3 + (1/5)x^4 + ... as an elementary function of x.

Without worrying too much about convergence:

Let

$f(x)=1+\frac{x}{2}+\frac{x^2}{3}+...$

then:

$\frac{d}{dx} xf(x)=1+x+x^2+... = \frac{1}{1-x}$

CB

(Also, you should have asked permission to post this in the maths challenge area of MHF)

**icemanfan** · January 14th 2012, 09:52 AM

Originally Posted by CaptainBlack

Without worrying too much about convergence:

Let

$f(x)=1+\frac{x}{2}+\frac{x^2}{3}+...$

then:

$\frac{d}{dx} xf(x)=1+x+x^2+... = \frac{1}{1-x}$

CB

It did occur to me to solve the problem that way, but then how do you determine what f(x) is?

$\frac{d}{dx} xf(x) = xf'(x) + f(x) = 1+x+x^2+... = \frac{1}{1-x}$

$xf'(x) + f(x) = \frac{1}{1-x}$

Not sure where you go from there. My solution is entirely different.

**uniquesailor** · January 15th 2012, 12:47 PM

**icemanfan** · January 16th 2012, 01:32 PM

Thanks, uniquesailor. That was easier than I made it. Actually, I realized my original solution was wrong, but I'm not sure why. This is what I did:

$e^x + \frac{e^x}{2x} + \frac{e^x}{3x^2} + \frac{e^x}{4x^3} + ... = e^xg(x)$

$\frac{d}{dx} e^x + \frac{e^x}{2x} + \frac{e^x}{3x^2} + \frac{e^x}{4x^3} =$

$e^x + \frac{e^x}{2x} - \frac{e^x}{x^2} + \frac{e^x}{3x^2} - \frac{e^x}{x^3} + \frac{e^x}{4x^3} - \frac{e^x}{x^4} ... =$

$e^x + \frac{e^x}{2x} + \frac{e^x}{3x^2} + \frac{e^x}{4x^3} + ... - \frac{e^x}{x^2} - \frac{e^x}{x^3} - \frac{e^x}{x^4} - ... =$

$e^x g(x) - \frac{e^x}{x^2}(1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ...)$

Hence

$\frac{d}{dx} e^x g(x) = e^x g(x) - \frac{e^x}{x^2} \cdot \frac{x}{x - 1}$

$e^x g(x) + e^x g'(x) = e^x g(x) - e^x \cdot \frac{1}{x(x - 1)}$

$e^x g'(x) = -e^x \cdot \frac{1}{x(x - 1)}$

$g'(x) = -\frac{1}{x(x - 1)}$

$g(x) = \ln (x) - \ln (x - 1) + C$

$f(x) = g(\frac{1}{x}) = \ln (\frac{1}{x}) - \ln (\frac{1 - x}{x}) + C$

$f(x) = -\ln (1 - x) + C$

Somehow I lost a factor of x in there...

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