What is $\sum_{i=1}^{r} (i^{-1/3})$?. Is there any general formula for $\sum_{i=1}^{r} (i^{-1/n})$ for any integer $n$?
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Using the Riemann zeta and Hurwitz zeta functions, $$\sum_{k=1}^rk^{p}=\zeta(p)-\zeta(p,r+1)$$ for any complex $p$. Thought probably not what you were looking for I assume? – Simply Beautiful Art Oct 09 '16 at 18:43
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The sum is $$r^{1-1/n}\frac1r\sum_{i=1}^r\left(\frac{i}r\right)^{-1/n}\sim r^{1-1/n}\int_0^1x^{-1/n}dx=\frac{n}{n-1}r^{(n-1)/n}$$ when $r\to\infty$ for $n>1$ fixed. – Did Oct 09 '16 at 18:47
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@Did Haha, that's not closed form, but good approximation. – Simply Beautiful Art Oct 09 '16 at 18:49
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2Yes, there exists no closed form. – Did Oct 09 '16 at 18:50