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If we know that the identity `sum_(k=1)^infty 1/k^2 = prod_(p quad "prime")...

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user7230927 | (Level 1) Salutatorian

Posted April 30, 2013 at 2:59 PM via web

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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`?

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mathsworkmusic | (Level 1) Educator

Posted April 30, 2013 at 5:48 PM (Answer #1)

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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).


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oldnick | (Level 1) Valedictorian

Posted May 1, 2013 at 4:36 AM (Answer #2)

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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.



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