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Opalg · #11 November 28th, 2008, 07:30 AM

Quote:

Originally Posted by davidmccormick

For the Weierstrass M-test what series do we use to show that $-\sum_{r=1}^{\infty}\frac{\ln r}{r^s}$ and $\sum_{r=1}^{\infty}\frac{(\ln r)^2}{r^s}$ respectively are uniformly convergent.

These series are uniformly convergent on the interval [a,∞), for any given a>1. In fact, $\frac{(\ln r)^k}{r^s}$ is decreasing as a function of s, because its derivative is $-\frac{(\ln r)^{k+1}}{r^s}$ . So its maximum value on the interval [a,∞) is its value at the left endpoint s=a, and we take $M_r = \frac{(\ln r)^k}{r^a}$ (where k = 1 or 2 as appropriate) in the M-test. The fact that $\sum_{r=1}^\infty \frac{(\ln r)^k}{r^a}$ converges (for a>1) then tells you that these series converge uniformly on that interval.

But note that uniform convergence on the interval [a,∞) for every a>1 does not imply uniform convergence on the interval (1,∞); and in fact the Riemann zeta series is not uniformly convergent on (1,∞).

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