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The series $\sum_{k=1}^\infty \frac{1}{k^\alpha}$ diverges if $\alpha\leq 1$. How can I estimate the divergent rate when $\alpha$ is given. For example, if $\alpha=1$, $\sum_{k=1}^n \frac{1}{k^\alpha}=O(\log n)$; if $\alpha=0$, $\sum_{k=1}^n \frac{1}{k^\alpha}=O(n)$.

Michael Hardy
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Wayson Kong
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    Hint: estimate this sum from below and from above by $\int_a^b x^{-\alpha}\mathrm dx$ where $a$ and $b$ have to be chosen in a proper way. – SBF Jan 05 '13 at 15:40

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You can use the Abel summation technique from here to derive the asymptotic Euler-Maclaurin formula given by $$\sum_{m=a}^b f(m) = \int_a^b f(x)dx + \dfrac{f(a) + f(b)}2 + \sum_{k=1}^{\infty} \dfrac{B_{2k}}{(2k)!}\left(f^{(2k-1)'}(b)-f^{(2k-1)'}(a)\right)$$ In your case, $f(x) = \dfrac1{x^{\alpha}}$, $a=1$, $b=n$. Hence, we get that $$\sum_{m=1}^n \dfrac1{m^{\alpha}} = \dfrac{n^{1-\alpha}-1}{1-\alpha} +\dfrac{1+n^{-\alpha}}2 + \sum_{k=1}^{\infty} \dfrac{B_{2k}}{(2k)!}(-\alpha)\cdots(-\alpha-2k+2)\left(\dfrac1{n^{\alpha+2k-1}} - 1\right)$$

Community
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