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I'd like to know if there is any approximation for the partial sum of zeta function $\sum_{i=1}^n \frac{1}{i^\beta}$ where $\beta$ is a real number less than or equal $1$,e.g., in the case $\beta = 1$, we have an approximation of $\ln (n)$.

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Harry
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  • If $Re(\beta) < 1$ : $\sum_{i=1}^n\frac1{i^\beta}= \int_1^{n+1}x^{-\beta} dx+\sum_{i=1}^n \int_i^{i+1}\int_i^x \beta t^{-\beta-1}dx = \frac{(n+1)^{1-\beta}}{1-\beta}+\sum_{i=1}^n O(n^{-\beta-1}) = \frac{n^{1-\beta}}{1-\beta}+O(n^{-\beta})$ for $Re(\beta) \ge 1$ it is the same with a constant $\zeta(\beta)$ – reuns Feb 02 '17 at 12:01

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In the same manner,

$$\sum_{i=1}^n\frac1{i^\beta}\sim\int_0^n\frac1{x^\beta}\ dx=\frac{n^{1-\beta}}{1-\beta}\quad;\beta<1$$

Better approximations may be done with the Euler-Maclaurin formula:

$$\sum_{i=1}^n\frac1{i^\beta}=\frac{n^{1-\beta}}{1-\beta}+\zeta(\beta)+\frac1{2n^\beta}-\frac\beta{12n^{\beta+1}}+\mathcal O\left(\frac1{n^{\beta+3}}\right)$$

if $\beta\ne1$, else,

$$\sum_{i=1}^n\frac1i=\ln(n)+\gamma+\frac1{2n}-\frac1{12n^2}+\mathcal O\left(\frac1{n^4}\right)$$

You can mess around and see how good this approximates as well as some other simpler approximations on this graph.

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