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 May 30th, 2013, 04:36 PM #1 Newbie   Joined: May 2013 Posts: 3 Thanks: 0 zeta function Hello, can someone help me with prooving this: $\sum_{i=2}^\infty (\zeta(i)-1)=1$ All what I did is, that I write it as a dobule sum but I don't know what next. :/ $\sum_{i=2}^\infty(\sum_{n=1}^\infty \frac{1}{n^i} -1))$ Thanks in advance.
May 30th, 2013, 11:22 PM   #2
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Re: zeta function

Hi !
invert the order of two sums
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 May 31st, 2013, 01:04 AM #3 Newbie   Joined: May 2013 Posts: 3 Thanks: 0 Re: zeta function Hello thanks for your answer, I have a couple of more question. I thought about it, but am I allowed to invert these two sums because they are not finite? and even if I am what happened with 1 in the brackets? and what's more I got this: $\sum_{i=2}^\infty(\sum_{n=1}^\infty \frac{1}{n^i} ))=\sum_{n=1}^\infty(\sum_{i=2}^\infty \frac{1}{n^i} ))=\sum_{n=1}^\infty(\frac{1}{1-\frac{1}{n}}-\frac{1}{n}-1)$ and this is not good because when n=1 there is a 0 in the denominator Am I doing something wrong? Thanks.
May 31st, 2013, 10:36 PM   #4
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Re: zeta function

You missed something. Look carefully at the important first step in the attachement to my preceding post.
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 June 1st, 2013, 05:10 AM #5 Newbie   Joined: May 2013 Posts: 3 Thanks: 0 Re: zeta function Aha, ok I got it now, thanks for your help.

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