C. Hinsley

24 July 2020

The Riemann zeta function (source: https://en.wikipedia.org/wiki/File:Riemann-Zeta-Func.png)

Riemann's zeta function is defined as follows:

$$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$

Expanding this series, we get:

$$\zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} + \dots$$

We can obtain the sum of all the terms with even bases of denominators by multiplying the series by its second term:

$$\frac{1}{2^s} \zeta(s) = \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + \frac{1}{8^s} + \frac{1}{10^s} + \dots$$

And, just as well, we can sum over all the terms with odd denominators by subtracting the above from the original series:

$$\zeta(s) - \frac{1}{2^s} \zeta(s) = \left(1-\frac{1}{2^s}\right)\zeta(s) = \frac{1}{1^s} + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \frac{1}{9^s} + \dots$$

Now we will assign this new series $\left(1-\frac{1}{2^s}\right)\zeta(s)$ to the name $Z$, so we don't have to write it out each time:

$$Z := \left(1-\frac{1}{2^s}\right)\zeta(s)$$

Note that in applying $\left(1-\frac{1}{2^s}\right)$ to the zeta function, we have effectively removed all terms for which the denominator's base is a multiple of two. The importance of this will soon become clear.

Let this point in our journey be referred to as Checkpoint $\alpha$ — we'll need to return here momentarily. ****Next, we look again to the second term of the series $Z$:

$$Z = \frac{1}{1^s} + \left[ \frac{1}{3^s} \right] + \frac{1}{5^s} + \frac{1}{7^s} + \frac{1}{9^s} + \dots$$

The second term is $\frac{1}{3^s}$. Again, we multiply this base by the series:

$$\frac{1}{3^s}Z = \frac{1}{3^s} + \frac{1}{9^s} + \frac{1}{15^s} + \frac{1}{21^s} + \frac{1}{27^s} + \dots$$

And just as before, we can subtract this from the whole series $Z$: