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Mathstud28 · #2 November 27th, 2008, 03:29 PM

Quote:

Originally Posted by davidmccormick

Consider the function

$f(s) = \sum_{r=1} ^ {\infty} \frac{1}{r^s}$ .
I have so far managed to show that the series converges for each $s\in\ (1,{\infty})$ and that this series defines a continuous function $f : (1,{\infty}) \rightarrow\mathbb{R}$ . I am however struggling to show that:

(i) $f$ is differentiable and that $f'(s) < 0$ for all $s \in\ (1,{\infty})$ .
(ii) $f$ is differentiable and that $f''(s) > 0$ for all $s \in\ (1,{\infty})$ .

Any help would be greatly appreciated.
thanks

Note that on the specified interval that $\zeta(x)$ converges uniformly on $(1,\infty)$ since those points are on the interior of its interval of convergence. Now note that $\frac{1}{r^x}$ is differentiable to $\frac{-x}{r^{x+1}}$ and that by the Weirstrass test or using the ratio/root test that this is uniformly convergent on $(0,\infty)$ we can conclude that

$\forall{x}\in(1,\infty)~\zeta'(x)=\sum_{n=1}^{\infty}\frac{-x}{r^{x+1}}$

Now it is obvious that $\forall{x}\in(1,\infty)~\frac{-x}{r^{x+1}}<0$ , so there is part i and for part two repeat a similar process of establishing, uniform convergence of $\sum_{n=1}^{\infty}\frac{-x}{r^{x+1}}$ on $(1,\infty)$ , uniform convergence of $\sum_{n=1}^{\infty}\frac{x(x+1)}{r^{x+2}}$ on $(1,\infty)$ , and the differentiability of $\frac{-x}{r^{x+1}}$ on $(1,\infty)$

The following users thank Mathstud28 for this useful post:
davidmccormick
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