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CaptainBlack · #8 November 28th, 2008, 04:21 AM

Quote:

Originally Posted by davidmccormick

But all I've done is differentiating the first term of the original series. I am pretty sure that the derivative exist and is equal to $\zeta'(s) = -\sum_{r=1}^{\infty}\frac{lnr}{r^s} = -\sum_{r=2}^{\infty}\frac{lnr}{r^s}$ . But I have not actually shown that $\zeta$ is differentiable.
Any thoughts.
thanks

It is sufficient to show that the sequence of parital sums:

$S_N(x)=\sum_{r=1}^N \frac{1}{r^x}$

in a neighbourhood of $x$ is uniformly convergent, and that $S'_N(x)$ is also uniformly convergent to conclude that:

$\frac{d \zeta}{dx}=\frac{d}{dx}\left[ \lim_{N \to \infty} S_N(x) \right] = \lim _{N \to \infty} \frac{dS_N(x)}{dx}$

The uniform convergence on the sequence of partial sums for the zeta function can be demonstrated fairly easily on any closed interval $[a,b],\ 1<a<b$ , and with a bit more trouble for the sequence of derivatives of the partial sums. Together these prove that the derivative of the zeta function can be found by term by term differentiation of the series for the zeta function on $(1,\infty).$

CB

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davidmccormick
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