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Mathstud28 · #4 November 27th, 2008, 07:05 PM

Quote:

Originally Posted by davidmccormick

I feel really stupid because your explanation is very clear but I don't follow....can you please explain why the derivative of $\frac{1}{r^x}$ is $\frac{-x}{r^{x+1}}$ because I make it out to be $-r^{-x}ln(r)$ . The second thing is the ratio test on $\frac{-x}{r^{x+1}}$ gives convergence, does that necessarily imply uniform convergence.
Sorry about this and thanks a lot

Because $\zeta(x)$ is a function of x so $\zeta'(x)=\frac{d}{dx}\sum_{n=1}^{\infty}\frac{1}{r^x}$ now since the appropriate conditions were met as I showed you $\frac{d}{dx}\sum_{n=1}^{\infty}\frac{1}{r^x}=\sum_{n=1}^{\infty}\frac{d}{dx}\frac{1}{r^x}=\sum_{n=1}^{\infty}\frac{-x}{r^{x+1}}$ . And to answer your other question, I mistyped, forgive me. Disregard the ratio/root test comment and stick the Weirstrass M-test.