While applying the Sieve of Eratosthenes to Zeta function we subtract $\frac{1}{2^s} \zeta(s)$ from $\zeta(s)$ to obtain
$$\zeta(s) - \frac{1}{2^s}\zeta(s) = \sum \frac{1}{n^s} - \sum \frac{1}{(2n)^s}$$
or
$$\left(1-\frac{1}{2^s} \right) \zeta(s) = 1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{6^s}+...$$
But when I do the subtraction I don't get the same result.
Since
$$\sum \frac{1}{n^s} - \sum \frac{1}{(2n)^s}$$
is
$$\sum \frac{1}{n^s} - \frac{1}{(2n)^s}$$
when I expand this, I get
$$\left(\frac{1}{1^s}-\frac{1}{2^s}\right) + \left(\frac{2^s}{4^s}-\frac{1}{4^s}\right) + \left(\frac{2^s}{6^s}-\frac{1}{6^s}\right)+...$$
or,
$$\frac{2^s - 1}{2^s} + \frac{2^s -1}{4^s} + \frac{2^s-1}{6^s}+...$$
Can you please exlain how I subtract these correctly?
First few terms:
1 1/2^s 1/3^s 1/4^s
1/2^s 1/4^s 1/6^s 1/8^s
At the end of the question, you have the series
$$\frac{2^s - 1}{2^s} + \frac{2^s -1}{4^s} + \frac{2^s-1}{6^s}+\cdots .$$
This is correct. It is merely not in the form you want.
The general term for $n = 1, 2, 3, \ldots$ is
$$ \frac{2^s - 1}{(2n)^s}. $$
Factor this as follows:
$$ \frac{2^s - 1}{(2n)^s} = \frac{2^s - 1}{2^s}\cdot \frac{1}{n^s}. $$
Now the series can be written like this:
$$ \frac{2^s - 1}{2^s}\cdot \frac{1}{1^s} + \frac{2^s - 1}{2^s}\cdot \frac{1}{2^s} + \frac{2^s - 1}{2^s}\cdot \frac{1}{3^s} + \cdots . $$
Now you just need to know that if you take a converging series and multiply every single term by the same factor, the sum of the series is multiplied by that same factor. In this case the converging series is $\dfrac{1}{1^s} + \dfrac{1}{2^s} + \dfrac{1}{3^s} + \cdots$ and the "same factor" is $\dfrac{2^s - 1}{2^s}.$ Therfore $$ \frac{2^s - 1}{2^s}\cdot \frac{1}{1^s} + \frac{2^s - 1}{2^s}\cdot \frac{1}{2^s} + \frac{2^s - 1}{2^s}\cdot \frac{1}{3^s} + \cdots = \frac{2^s - 1}{2^s}\left(\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \right). $$
Finally, recognize that $$\frac{2^s - 1}{2^s} = 1 - \frac{1}{2^s}$$ so that you can write the result the same way as in the paper.
For the part where we want to show that the difference of the two series is equal to the series consisting of only the "odd" terms of $\zeta(s),$ because the series $\dfrac{2^s - 1}{2^s} + \dfrac{2^s -1}{4^s} + \dfrac{2^s-1}{6^s}+\cdots$ is absolutely convergent, you can change it to
$$\frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{2^s} - \frac{1}{4^s} + \frac{1}{3^s} - \frac{1}{6^s}+\cdots$$
and then, because it's still absolutely convergent, you can rearrange the order of the terms any way you want, including shifting all the negative terms by increasing number of places to the right so that the sequence of signs is $++-++-++-\cdots$ instead of $+-+-+-+-\cdots.$ That is, it is perfectly OK to write
$$\frac{1}{1^s} + \frac{1}{2^s} - \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} - \frac{1}{4^s} + \frac{1}{5^s}+\cdots$$
and so forth. Finally, you can combine the terms three at a time so that the last two terms in each group of three cancel.
I don't think this would necessarily be OK if it involved any conditionally convergent series, but we have absolute convergence here.