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

  • $\begingroup$ 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


(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.

  • $\begingroup$ 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

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