3
$\begingroup$

The Riemann zeta function is defined on the $Re z> 1$ by $$\zeta(z)=\sum_{n=1}^\infty \frac{1}{n^z}$$

(i) show that for $Re z> 1$, we have $$(1-2^{1-z})\zeta(z)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}$$%

(ii) show that $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}$$ is an analytic function on $Re z> 0$

Thoughts thus far:

(i) Since $$2^z=2^{Rez+iImz}=2^{Rez}2^{iImz}=2^{Rez}[\cos(Imz\ln2)+i\sin(Imz\ln2)]$$ we obtain $$(1-2^{1-z})\zeta(z)=\sum_{n=1}^\infty \frac{1}{n^z}-\sum_{n=1}^\infty \frac{2^{1-z}}{n^z}=\sum_{n=1}^\infty \frac{1}{n^z}-\sum_{n=1}^\infty \frac{2}{2^{Rez}[\cos(Imz\ln2)+i\sin(Imz\ln2)]n^z}=$$ (by multiplying by the conjugate) $$\sum_{n=1}^\infty \frac{1}{n^z}-\sum_{n=1}^\infty \frac{2(\cos(Imz\ln2)-i\sin(Imz\ln2))}{2^{Rez}[\cos^2(Imz\ln2)+\sin^2(Imz\ln2)]n^z}=$$ (since $\sin^2\theta+\cos^2\theta=1$) $$\sum_{n=1}^\infty \frac{1}{n^z}-\sum_{n=1}^\infty \frac{2(\cos(Imz\ln2)-i\sin(Imz\ln2))}{2^{Rez}n^z}=\sum_{n=1}^\infty \frac{2^{Rez}-2(\cos(Imz\ln2)-i\sin(Imz\ln2))}{2^{Rez}n^z}$$, at which point I get stuck. I am uncertain whether unraveling $2^z$ in the manner that I did was fruitful or perhaps, as usual, I am missing something berry basic.

(ii) Since we want to show analyticity, we may show that the power series converges. Using the logic as above, we know that $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^{Rez}[\cos(Imz\ln n)+i\sin(Imz\ln n)]}=\sum_{n=1}^\infty \frac{(-1)^{n+1}[\cos(Imz\ln n)-i\sin(Imz\ln n)]}{n^{Rez}}$$ Since $\cos(Imz\ln n)-i\sin(Imz\ln n)$ represents oscillations around the unit circle we may clearly see that $$\lim_{n \to \infty}|\frac{(-1)^{n+1}\cos(Imz\ln n)-i\sin(Imz\ln n)}{n^{Rez}}|=\lim_{n \to \infty}|\frac{1}{n^{Rez}}|=0\iff Rez>0$$ and hence the power series converges if and only if Rez>0. Does this appear to be a valid proof for analyticity?

Thanks in advance for any help that you may provide

$\endgroup$
4
  • $\begingroup$ hints for (i) separate the even and odd parts (observe that $(2n)^z=2^z\,n^z$) and please keep $z$ as a whole! :-) $\endgroup$ Oct 27, 2012 at 9:43
  • $\begingroup$ for (ii) here is a proof by Rob Johnson. $\endgroup$ Oct 27, 2012 at 9:49
  • $\begingroup$ @RaymondManzoni Thank you for your response. The proof that Johnson did seems a bit involved, which makes me believe that my analysis above for part (ii) is flawed somehow. For part (ii), what is incorrect about the method that I employed? $\endgroup$
    – ABC Bach
    Oct 27, 2012 at 10:07
  • $\begingroup$ Rob Johnson's proof is interesting to prove convergence for $\Re(s)>0$ (the $0$ limit you got is insufficient). But for your question $\Re(s)>1\ $ is merely supposed so that absolute convergence should be enough (as with $\zeta$ itself) and the integral test should allow to conclude (see too Dirichlet series) $\endgroup$ Oct 27, 2012 at 12:29

1 Answer 1

4
$\begingroup$

(i) For $\Re(z)> 1$ we have : $$ \begin{align} \zeta(z)&=\sum_{n=1}^\infty \frac 1{n^z}\\ &=\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^z}+2\sum_{n=1}^\infty \frac 1{(2n)^z}\\ &=\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^z}+2^{1-z}\,\zeta(z)\\ \end{align} $$ and the result :

$$(1-2^{1-z})\zeta(z)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}$$


(ii) For $\Re(z)> 0$ :

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}=\sum_{n=1}^\infty \frac 1{(2n-1)^z} -\frac 1{(2n)^z}$$

You may use the 'mean value theorem' applied to $f(x)=x^{-z}$ to prove the existence of a real '$c$' verifying $\,2n-1\le c \le 2n\,$ and such that : $$f'(c)=\frac{f(2n)-f(2n-1)}1$$

Since $f'(c)=-z\,c^{-z-1}\,$ we have : $$f(2n-1)-f(2n)=\frac z{c^{z+1}}$$ getting the upper bound : $$|f(2n-1)-f(2n)|\le \left|\frac z{(2n-1)^{z+1}}\right|$$

and the majoration of our alternate series : $$\left|\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^z}\right|\ \le\ \sum_{n=1}^\infty \;\left|\frac z{(2n-1)^{z+1}}\right|$$

The right part is simply $\ \displaystyle f(x)=\sum_{n=1}^\infty \frac {|z|}{(2n-1)^{x+1}}\ $ with $\ x:=\Re(z)$.

Using the integral test with the observation that $\displaystyle\int_1^\infty \frac {dn}{\,(2n-1)^{x+1}}=\left[\frac {-1}{2x\,(2n-1)^x}\right]_{n=1}^\infty=\frac 1{2x} $ for $x > 0$ we conclude that the alternate series is convergent for $\Re(z)> 0$.

For more about Dirichlet series see this Wikipedia link.

$\endgroup$
5
  • $\begingroup$ Thank you for the in-depth explanation. Your result appears to depend only upon $\Re(z)> 1$. Why would the question then ask for us to prove that the series is analytic for the right half of the real plane? $\endgroup$
    – ABC Bach
    Oct 28, 2012 at 9:18
  • $\begingroup$ @ABCBach: Is that the question (iii) or more because you didn't specify this in (i) nor (ii) ? I think that indeed the point of the alternate series is to get convergence in the right half plane (so my earlier comments). $\endgroup$ Oct 28, 2012 at 9:24
  • $\begingroup$ That is because I originally posted the problem incorrectly (part (ii) was for $\Re(z)>0$). I sincerely apologize for the inconvenience. This is fixed now. $\endgroup$
    – ABC Bach
    Oct 28, 2012 at 9:30
  • $\begingroup$ @ABCBach: I'll update my answer with another proof. $\endgroup$ Oct 28, 2012 at 9:32
  • $\begingroup$ Thank you. Your help is very much appreciated. $\endgroup$
    – ABC Bach
    Oct 28, 2012 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.