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How would you show that the Riemann Zeta function, $\zeta(s) < 0$ for $s \in (0,1)$?

So far I have that along the critical strip

\begin{align} \zeta(s) &= \left(\frac{2^{s-1}}{2^{s-1}-1}\right)\phi(s)\\ &= \left(\frac{1}{1-2^{1-s}}\right)\phi(s)\\ &= \left(\frac{1}{1-2^{1-s}}\right)\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s} \end{align}

Where $\phi(s)$ is Euler's alternating zeta function. (which converges for $s > 0$) How would you show that this is always negative when $s \in (0,1)$?

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3 Answers 3

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If $s\in(0,1)$ then $\frac{1}{1-2^{1-s}}<0$, while $$\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n^s}$$ is positive due to Leibniz' criterion.

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    $\begingroup$ Jack, your answers are always the best, thank you so much! $\endgroup$
    – Pablo
    Commented Aug 25, 2014 at 8:27
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    $\begingroup$ (+1) This is the way I usually show this, but your answer prodded me to bring out slightly heavier artillery. $\endgroup$
    – robjohn
    Commented Aug 25, 2014 at 9:29
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Note that $$ \varphi(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=\sum_{k=1}^\infty \left(\frac{1}{(2k-1)^s}-\frac{1}{(2k)^s}\right)>0, $$ as $$ \frac{1}{(2k-1)^s}>\frac{1}{(2k)^s}. $$

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Here is an approach using the integral for $\zeta(z)\Gamma(z)$ and a simple analytic continuation.

Waxing complex: For $\mathrm{Re}(z)\gt1$, we have $$ \begin{align} \zeta(z)\Gamma(z) &=\int_0^\infty\frac{t^{z-1}}{e^t-1}\mathrm{d}t\\ &=\frac1{z-1}\int_0^\infty\frac{t}{e^t-1}\mathrm{d}t^{z-1}\\ &=\frac1{z-1}\int_0^\infty t^{z-1}\frac{1+(t-1)e^t}{(e^t-1)^2}\mathrm{d}t\tag{1} \end{align} $$ Since $\frac{\mathrm{d}}{\mathrm{d}t}\left[1+(t-1)e^t\right]=te^t\ge0$ for $t\ge0$, the integrand is positive. The integral in $(1)$ also converges for $\mathrm{Re}(z)\gt0$ and is analytic. Therefore, $(1)$ represents the analytic continuation of $\zeta(z)\Gamma(z)$ to $\mathrm{Re}(z)\gt0$.

Back to the real world: Since $\Gamma(z)\gt0$ for $z\gt0$, $(1)$ says that $\zeta(z)\lt0$ for $0\lt z\lt1$.

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  • $\begingroup$ (+1) The next volume of "Mathematics made difficult" will be yours :D $\endgroup$ Commented Aug 25, 2014 at 9:35
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    $\begingroup$ Thanks. However, showing that $\zeta(s)=\frac1{1-2^{1-s}}\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}$ for $0\lt s\lt1$, requires some analytic continuation. One approach is detailed in the beginning of this answer. $\endgroup$
    – robjohn
    Commented Aug 25, 2014 at 9:54

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