3
$\begingroup$

Wikipedia lists a series expansion for $\zeta(s)$ here. How is the Dirichlet series below derived? I apologize in advance if this is a very simple question, I don't know much about Dirichlet series.

Wikipedia summation formula

$$\zeta(s)=\frac{1}{s-1}\sum_{n=1}^\infty \left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right)$$ which converges for $\Re(s) > 0$

$\endgroup$

1 Answer 1

4
$\begingroup$

I'm not sure why Wikipedia has it in that form, but here's a plausible explanation of how one could derive an equivalent form. The first question to ponder is how $\sum_{k=1}^\infty n^{-s}$ grows if ${\rm Re}(s)<1$. We get

$$\sum_{k=1}^n \frac{1}{k^s} =n^{1-s}\sum_{k=1}^n \frac{1}{n}\left(\frac{k}{n}\right)^{-s}\sim n^{1-s}\int_0^1 x^{-s}{\rm d}x =\frac{n^{1-s}}{1-s}$$

for ${\rm Re}(1-s)>0\iff {\rm Re}(s)<1$. Heuristically we find that

$$\sum_{k=1}^n \frac{1}{k^s} \approx \frac{n^{1-s}}{1-s}+\zeta(s) $$

for ${\rm Re}(s)>0,\ne1$ by checking the cases $0<{\rm Re}(s)<1$ and ${\rm Re}(s)>1$ separately. Since $\zeta(s)$ has a pole at $s=1$ let's look at the function $(s-1)\zeta(s)$ instead. It is heuristically the limit of the terms

$$ c_n=(s-1)\sum_{k=1}^n\frac{1}{k^s}+n^{1-s}. $$

We check the forward differences are

$$\begin{array}{ll} c_{n+1}-c_n & \displaystyle =\frac{s-1}{(n+1)^s}+\frac{n+1}{(n+1)^s}-\frac{n}{n^s} \\ & \displaystyle =\frac{s}{(n+1)^s}+n\left(\frac{1}{(n+1)^s}-\frac{1}{n^s}\right). \end{array} $$

Therefore, adding $c_1+(c_2-c_1)+\cdots+(c_{n+1}-c_n)$ (note $c_1=s$) we get

$$\begin{array}{ll} c_{n+1} & \displaystyle =s+\sum_{k=1}^n \left[\frac{s}{(k+1)^s}+k\left(\frac{1}{(k+1)^s}-\frac{1}{k^s}\right)\right] \\ & \displaystyle = \frac{s}{(n+1)^s}+\sum_{k=1}^n \left[ \frac{s}{k^s}+k\left(\frac{1}{(k+1)^s}-\frac{1}{k^s}\right)\right]. \end{array}$$

The term $\displaystyle \frac{s}{(n+1)^s}$ out in front $\to0$ as $n\to\infty$ so delete. Take the limit to get

$$(s-1)\zeta(s)=\sum_{k=1}^\infty \left[ \frac{s}{k^s}+k\left(\frac{1}{(k+1)^s}-\frac{1}{k^s}\right) \right] $$

which is the desired form.

$\endgroup$
3
  • $\begingroup$ Thank You, excellent answer. Struggling with the third step at the beginning which forms n^(1-s)*\int_0^1*x^(-s)dx. I don't see how this part was formed, I will think about this some more. Thank You very much. $\endgroup$
    – Axion004
    Sep 23, 2015 at 3:45
  • 1
    $\begingroup$ @Axion004 $\displaystyle\sum_{k=1}^n \frac{1}{n}\left(\frac{k}{n}\right)^{-s}$ is the Riemann sum for $\displaystyle\int_0^1 x^{-s}{\rm d}x$. (Standard trick BTW.) $\endgroup$
    – whacka
    Sep 23, 2015 at 4:53
  • $\begingroup$ @whacka: Could you show me why $\sum_{k=1}^n \frac{1}{n}\left(\frac{k}{n}\right)^{-s}=\displaystyle\int_0^1 x^{-s}{\rm d}x$? Any reference is helpful. Thanks $\endgroup$
    – mike
    Jun 10, 2016 at 10:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .