Can I simplify the sum: $\zeta(1) +\zeta(2) +...+\zeta(n)$ . Whereas $\zeta$ denotes the zeta function. I want to find the limit involving this sum but need it simplified. Thanks.
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$\zeta(1)$ is equal to complex infinity. So perhaps you mean to start with $\zeta(2)$? – Caleb Stanford Nov 27 '13 at 17:16
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This question was already asked several times: http://math.stackexchange.com/questions/576266/finite-sum-of-zita-functions, and http://math.stackexchange.com/questions/576769/finite-sum-of-riemann-function. – Dietrich Burde Nov 27 '13 at 19:54
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Assuming you meant $\zeta(2) + \zeta(3) + \cdots + \zeta(n)$.
\begin{align*} \sum_{i=2}^n \zeta(i) &= \sum_{i=2}^n \sum_{k=1}^\infty \frac{1}{k^{i}} \\ &\stackrel{?}{=} \sum_{k=1}^\infty \sum_{i=2}^n \frac{1}{k^{i}} \\ &= \sum_{k=1}^\infty \frac{(1/k)^2 - (1/k)^{n+1}}{1 - (1/k)} \end{align*}
You need to justify the $\stackrel{?}{=}$ step. Then take the limit as $n \to \infty$.
Caleb Stanford
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