If we know that the identity

`sum_(k=1)^infty 1/k^2 = prod_(p quad "prime") 1/(1-p^(-s))`

holds for `s in {2,3,4}`

does this identity hold for any other values of `s`?

### 2 Answers | Add Yours

Euler's product formula states that

`prod_(p quad "prime") 1/(1-p^(-s)) = sum_(k=1)^infty 1/k^s`

So if we have that `sum_(k=1)^infty 1/k^2 = prod_(p quad "prime")1/(1-p^(-s))`

for `s in {2,3,4,...}` then we must also have that

`sum_(k=1)^infty 1/k^2 = sum_(k=1)^infty 1/k^s` for `s in {2,3,4,...}`

This implies that `k^2 = k^3 = k^4` and that `1/k^2` is defined.

The only value of `k` for which this could be true is `k=1`.

**Since `1^2 = 1^s` for any `s in RR`, `s` can take any value on the real numberline (or indeed could be complex).**

let it be:

`xi (s)=sum_(k=1)^oo 1/k^s` `=prod_(p) 1/(1-p^s)`

If it were, by absurd: `prod_(p) 1/(1-p^s)=sum_(k=1)^oo 1/k^2` `AA` s>2 we can get:

`sum_(s=2)^oo xi (s)=sum_(s=2)^oo sum_(k=1)^oo 1/k^s=` `sum_(k=1)^oo 1/k^2+ sum_(s>2)^oo sum_(k=0)^oo 1/k^s =`

`=xi (s) +sum_(s>2)^oo sum_(k=1)^oo 1/k^s=` `sum_(s>2)^oo prod_(p) 1/(1-p^s)` (1)

right side of (1) we can write as:

`sum_(s>2)^oo prod_(p) 1/(1-p^s)=sum_(s>2)^oo prod_(p) sum_(h=0)^(-oo)(p^s)^h =` (2)

For distribution rule of moltiplication the rigth side, compute all integer number as h runs all integer along. So that:

`sum_(s>2)^oo prod_(p) 1/(1-p^s)=` `sum_(k=1)^oo sum_(s>2)^oo 1/k^s` `` `=sum_(k=1)^oo k/(k-1)` (3)

Substituing this in (1) we have:

`xi (s)+sum_(s>2)^oo sum_(k=1)^oo 1/k^s=sum_(k=1)^oo k/(k-1)` (4)

Now the series at right side of (4) diverges, while both `xi (s)` than the others series on the left side converges, thus the absurd.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes