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 November 10th, 2013, 09:03 AM #1 Newbie   Joined: Oct 2013 Posts: 18 Thanks: 0 1st year math - Riemann zeta function & series the riemann zeta function is defined for p > 1 as $\zeta(p) := \displaystyle \sum_{n=1}^\infty \dfrac{1}{n^p}$ prove $\displaystyle \sum_{p=2}^\infty [\zeta(p) - 1] = 1$ by considering $\displaystyle \right | 1 - \sum_{p=2}^P \sum_{n=2}^N\dfrac{1}{n^p} \left | < \epsilon=$ so my thoughts: as $\displaystyle \sum_{p=2}^\infty [\zeta(p) - 1] = \sum_{p=2}^\infty \sum_{n=2}^\infty \dfrac{1}{n^p}$ then we just use the definition of convergence to prove that the partial sum converges to 1 and we're done however I'm having troubles trying to show that for some epsilon there exists a real number H s.t. P>H and N>H implies$\displaystyle \right | 1 - \sum_{p=2}^P \sum_{n=2}^N\dfrac{1}{n^p} \left | < \epsilon=$ I've tried using geometric series for the inside sum and then trying to rearrange for P or N but to no avail - any help please
 November 10th, 2013, 11:42 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115 Re: 1st year math - Riemann zeta function & series Have you noticed that: $\frac{1}{N(N-1)}= \frac{1}{N-1} - \frac{1}{N} ?$
November 10th, 2013, 12:28 PM   #3
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Re: 1st year math - Riemann zeta function & series

Quote:
 Originally Posted by Pero Have you noticed that: $\frac{1}{N(N-1)}= \frac{1}{N-1} - \frac{1}{N} ?$
I have not reached a stage where I can use that - could you please show me where?

 November 10th, 2013, 12:42 PM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115 Re: 1st year math - Riemann zeta function & series If you re-arrange into geometric series: $(\frac{1}{2^2} + \frac{1}{2^3} + ... \frac{1}{2^P}) + (\frac{1}{3^2} + \frac{1}{3^3} + ... \frac{1}{3^P}) + ... + (\frac{1}{N^2} + \frac{1}{N^3} + ... \frac{1}{N^P})$ Then, each sums to: $\frac{1}{n(n-1)}= \frac{1}{n-1} - \frac{1}{n} \ for \ n = 2, 3 ... N \ as \ P \rightarrow \infty$ Does that help?
November 10th, 2013, 12:58 PM   #5
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Re: 1st year math - Riemann zeta function & series

Quote:
 Originally Posted by Pero If you re-arrange into geometric series: $(\frac{1}{2^2} + \frac{1}{2^3} + ... \frac{1}{2^P}) + (\frac{1}{3^2} + \frac{1}{3^3} + ... \frac{1}{3^P}) + ... + (\frac{1}{N^2} + \frac{1}{N^3} + ... \frac{1}{N^P})$ Then, each sums to: $\frac{1}{n(n-1)}= \frac{1}{n-1} - \frac{1}{n} \ for \ n = 2, 3 ... N \ as \ P \rightarrow \infty$ Does that help?
I understand what you're doing but I'm considering the partial sums, why are you taking P to infinity?

November 10th, 2013, 01:20 PM   #6
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Re: 1st year math - Riemann zeta function & series

Quote:
 Originally Posted by lun123 [ why are you taking P to infinity?
Where else would you like to take it? N's got to get there too!

 November 11th, 2013, 10:21 AM #7 Newbie   Joined: Oct 2013 Posts: 18 Thanks: 0 Re: 1st year math - Riemann zeta function & series I'm sorry but I think you misunderstood what I was asking: I'm asking, how do I show that for all $\epsilon > 0$ there exists a real number H s.t. P>H and N>H implies $|1-\displaystyle\sum_{p=2}^P\sum_{n=2}^N(\dfrac{1}{n^ p}) | < \epsilon=$

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