Simple Riemann zeta function algebra help

Hi

It's just that last step I'm not getting, so you have:
[1 / Kz] - [1 / (2K)z]
= [ (2K)z - Kz ] / [(2K)z * Kz]
= [ (2)z - 1 ] / [(2K)z*]
Then what?
Thanks

Mark44
Mentor
Hi
View attachment 80246
It's just that last step I'm not getting, so you have:
[1 / Kz] - [1 / (2K)z]
= [ (2K)z - Kz ] / [(2K)z * Kz]
= [ (2)z - 1 ] / [(2K)z*]
Then what?
Thanks
Expand the first series, and you get ##\frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + ... + \frac{1}{n^k} ... ##
Expand the second series, and you get ##\frac{1}{2^k} + \frac{1}{4^k} + ... + \frac{1}{(2n)^k} ... ##
If you subtract the second series from the first, term by term, what do you get?

Expand the first series, and you get ##\frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + ... + \frac{1}{n^k} ... ##
Expand the second series, and you get ##\frac{1}{2^k} + \frac{1}{4^k} + ... + \frac{1}{(2n)^k} ... ##
If you subtract the second series from the first, term by term, what do you get?
Sorry, I'm not sure, I'm too rusty.

Mark44
Mentor
Can you see that every term in the second series is also present in the first series?

Can you see that every term in the second series is also present in the first series?
Ah, I've been trying to use factorisation and cancellation.
So when you minus S1 - S2 it leaves only 1/1k
but how does that explain the final line of working they got for the sum?

Mark44
Mentor
Ah, I've been trying to use factorisation and cancellation.
So when you minus S1 - S2 it leaves only 1/1k
No, that's not right. You still end up with a series with an infinite number of terms. Every term in the 2nd series is in the first series, but not every term in the 1st series is in the second. Expand both series like I did above, but with more terms, and then do the subtraction.
tim9000 said:
but how does that explain the final line of working they got for the sum?

mathman