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Opalg · #7 November 28th, 2008, 04:19 AM

Quote:

Originally Posted by davidmccormick

I am pretty sure that the derivative exist and is equal to $\zeta'(s) = -\sum_{r=1}^{\infty}\frac{\ln r}{r^s} = -\sum_{r=2}^{\infty}\frac{\ln r}{r^s}$ .

The series $-\sum_{r=1}^{\infty}\frac{\ln r}{r^s}$ converges uniformly in any interval [a,∞), where a>1 (by the Weierstrass M-test). So you can integrate this series term by term (getting $\zeta(s)$ ), and by the fundamental theorem of calculus the integrated series will have the required derivative, which is clearly negative.

The same argument, repeated, will show that the second derivative is $\sum_{r=1}^{\infty}\frac{(\ln r)^2}{r^s}$ , which is clearly positive.

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