<s>[INST] Show in between steps in this Riemann zeta function equivalence/reduciton 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.
 [/INST] 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...)
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