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Moo · #5 November 28th, 2008, 12:37 AM

Quote:

Originally Posted by Mathstud28

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.

As you said, it is a function of x.

$\frac{d}{dx} \left(\frac{1}{r^x} \right)=\frac{d}{dx} \left(e^{-x \ln(r)}\right)=-\ln(r) \cdot \frac{1}{r^x}$

So davidmccormick, you are correct.

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