Math Table or Method of Solving an infinite sum of reciprocal powers


by pasqualrivera
Tags: infinite sums
pasqualrivera
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#1
Sep28-10, 06:26 PM
P: 2
Is anybody aware of how to solve the following infinite sum:

[tex]\sum\frac{1}{n^2}[/tex] for all positive odd integers?



Is this the sort of thing you just look up in a math table or solve?

If math table, do I need a "sum of reciprocal powers" table or a "riemann zeta function" table?

If solve - how?

I know the answer is pi^2 / 8 but I haven't a clue how to calculate that and cannot find a math table with the appropriate functions.
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phyzguy
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#2
Sep28-10, 07:24 PM
P: 2,080
This is the Riemann zeta function Zeta(2), by definition. It's actually equal to pi^2/6, not pi^2/8, and Euler found a clever way of showing this, which is detailed here:

http://en.wikipedia.org/wiki/Basel_problem
Dick
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#3
Sep28-10, 10:04 PM
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Quote Quote by phyzguy View Post
This is the Riemann zeta function Zeta(2), by definition. It's actually equal to pi^2/6, not pi^2/8, and Euler found a clever way of showing this, which is detailed here:

http://en.wikipedia.org/wiki/Basel_problem
No, it's pi^2/8. Not zeta(2). The OP asked for the sum over odd integers. Split the sum over all integers into the sum over even integers and the sum over odd and equate it to zeta(2). Now find a way to relate the sum over even integers to zeta(2).

phyzguy
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#4
Sep29-10, 07:25 AM
P: 2,080

Math Table or Method of Solving an infinite sum of reciprocal powers


Sorry, I missed the "odd". You're right.
pasqualrivera
pasqualrivera is offline
#5
Sep29-10, 03:35 PM
P: 2
That makes sense, since the even series is not hard to find. I was thinking about this earlier today and haven't done the calculation yet, but I'm glad to have some reinforcement on that path.

Now I get to have fun learning the even series.

thx


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