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.
It did occur to me to solve the problem that way, but then how do you determine what f(x) is?
$\displaystyle \frac{d}{dx} xf(x) = xf'(x) + f(x) = 1+x+x^2+... = \frac{1}{1-x}$
$\displaystyle xf'(x) + f(x) = \frac{1}{1-x}$
Not sure where you go from there. My solution is entirely different.
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:
$\displaystyle e^x + \frac{e^x}{2x} + \frac{e^x}{3x^2} + \frac{e^x}{4x^3} + ... = e^xg(x)$
$\displaystyle \frac{d}{dx} e^x + \frac{e^x}{2x} + \frac{e^x}{3x^2} + \frac{e^x}{4x^3} = $
$\displaystyle 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} ... = $
$\displaystyle 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} - ... = $
$\displaystyle e^x g(x) - \frac{e^x}{x^2}(1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ...)$
Hence
$\displaystyle \frac{d}{dx} e^x g(x) = e^x g(x) - \frac{e^x}{x^2} \cdot \frac{x}{x - 1}$
$\displaystyle e^x g(x) + e^x g'(x) = e^x g(x) - e^x \cdot \frac{1}{x(x - 1)}$
$\displaystyle e^x g'(x) = -e^x \cdot \frac{1}{x(x - 1)}$
$\displaystyle g'(x) = -\frac{1}{x(x - 1)}$
$\displaystyle g(x) = \ln (x) - \ln (x - 1) + C $
$\displaystyle f(x) = g(\frac{1}{x}) = \ln (\frac{1}{x}) - \ln (\frac{1 - x}{x}) + C$
$\displaystyle f(x) = -\ln (1 - x) + C$
Somehow I lost a factor of x in there...