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In the answer chosen by the OP in this question I had trouble understanding the steps taken to get the equivalences/reduce the zeta function into another one. Can somebody show me the steps to go from one step to the next in this:

$$\begin{align} \zeta(z)&=\frac 1{1^z}+\frac 1{2^z}+\frac 1{3^z}+\frac 1{4^z}+\frac 1{5^z}+\cdots\\ &=\frac 1{1^z}-\frac 1{2^z}+\frac 1{3^z}-\frac 1{4^z}+\frac 1{5^z}+\cdots+\frac 2{2^z}+\frac 2{4^z}+\cdots\\ &=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+\frac 2{2^z}\left(1+\frac 1{2^z}+\frac 1{3^z}\cdots\right)\\ \zeta(z)&=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+2^{1-z}\zeta(z)\\ \end{align}$$

By step, I mean to go from one equals sign to the next.

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zerosofthezeta
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  • You are right in your appreciation of @Raymond Manzoni. You may be interested in a very readable and beautiful book: Stopple's Primer on Analytic Number Theory. –  Sep 11 '13 at 21:25
  • @Andrew Thanks so much for the reference! I will definitely buy that book. I appreciate it! – zerosofthezeta Sep 12 '13 at 07:20
  • By the way, you may be interested why the first value of $\zeta$ on the critical line $1/2 + it$ where $t = 0$ is $negative$, i.e., takes the negative square root by isolating $\zeta(s)$ on the LHS and letting $s = 1/2$ –  Sep 13 '13 at 20:52

1 Answers1

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I have added three intermediate states.

\begin{align} \zeta(z)&=\frac 1{1^z}+\frac 1{2^z}+\frac 1{3^z}+\frac 1{4^z}+\frac 1{5^z}+\frac 1{6^z}+\frac 1{7^z}+\cdots\qquad\qquad\qquad(1)\\ &=\frac 1{1^z}+\left(-\frac 1{2^z}+\frac 2{2^z}\right)+\frac 1{3^z}+\left(-\frac 1{4^z}+\frac 2{4^z}\right)+\frac 1{5^z}+\left(-\frac 1{6^z}+\frac 2{6^z}\right)+\frac 1{7^z}+\cdots\\ &=\left(\frac 1{1^z}-\frac 1{2^z}+\frac 1{3^z}-\frac 1{4^z}+\frac 1{5^z}-\frac 1{6^z}+\frac 1{7^z}+\cdots\right)+\frac 2{2^z}+\frac 2{4^z}+\frac 2{6^z}+\cdots\\ &=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+\frac 2{2^z\,1^z}+\frac 2{2^z\,2^z}+\frac 2{2^z\,3^z}+\cdots\\ &=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+\frac 2{2^z}\left(\frac 1{1^z}+\frac 1{2^z}+\frac 1{3^z}+\cdots\right)\\ \zeta(z)&=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+2^{1-z}\;\zeta(z)\qquad\text{from}\ (1)\ \,\text{and since}\ \,2^{1-z}=\frac {2^1}{2^z}\\ \end{align}

If this is not cleared please let me know (I'll try a more classical $\sum$ formulation...)

Raymond Manzoni
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  • Beautifully explained! Thank you so much! Are you an expert on the zeta function? Is it okay if I contact you if I have any questions or tag you in my questions? – zerosofthezeta Sep 11 '13 at 17:44
  • Thanks for all that @zerosofthezeta ! I am not really an expert even if I played often with $\zeta$ since $1998$. I think that putting names in a question is not really politically correct here and, worse, would make you miss answers from excellent contributors (zeta is rather well known). Anyway the 'riemann-zeta' tag should bring your question to my attention (or a comment here or under any of my answers with a link to your question). Great explorations ! – Raymond Manzoni Sep 11 '13 at 21:01
  • Hey Raymond, I tried working through your steps just to get a better understanding and ran into trouble. When you factor out $\frac 2{2^z}$ shouldn't the sum inside the parentheses right next to it NOT include the power of z on each term because of the power law: $(a^m)(a^m)=(a^m)^2=(a^{2m})$? – zerosofthezeta Sep 13 '13 at 05:51
  • Also, in the fourth equal sign down on the right side are the expressions with the 2's correct, because doesn't $\frac2{2^z 2^z} = \frac2{4^{2z}}$ ? – zerosofthezeta Sep 13 '13 at 05:54
  • @zerosofthezeta: Remember that $;a^z,b^z=(a,b)^z;$ so that $;2^z,2^z=(2\cdot 2)^z;$. In this specific case you may use too $;a^y,a^z=a^{y+z};$ getting $;2^z,2^z=2^{z+z}=2^{2z}=\left(2^2\right)^z;$ (in the first case we multiply the bases, in the second we add the exponents but we can't do both at once). Concerning the previous question I merely rewrote $,\frac 2{2^z,3^z}=\frac 2{2^z}\frac 1{3^z};$ and factorized the common terms $\frac 2{2^z}$. – Raymond Manzoni Sep 13 '13 at 12:01
  • I see...thanks again! – zerosofthezeta Sep 14 '13 at 06:48