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We all know that $$ \sum_{k=1}^n \frac{1}{k} \sim \log(n).$$

What is the analagous result for $\sum_{k=1}^n k^{-1/2}$ ?

Many thanks for your help.

Michael Hardy
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Frank
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  • It should be $O(\sqrt{n})$, but deriving the leading order behavior and possibly the next order is more interesting. – Ron Gordon Mar 11 '14 at 21:49
  • I have now edited the post. – Frank Mar 11 '14 at 21:55
  • You can bound $\sum_k f(k)$ by $\int_{k=?}^\infty f(k) dk$, as long as $f(.)$ is non-increasing/non-decreasing. The key is to interpret it as a Riemann sum of $f(k)$. – Hoda Mar 11 '14 at 21:58
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    The complete asymptotic expansion can be found at this MSE link. – Marko Riedel Mar 11 '14 at 22:08
  • I changed \textrm{~}\hspace{3mm} to \sim, and log(n) to \log(n). The takes care of the spacing, makes the tilde match the font, and prevents italicization of $\log$. It's standard usage. – Michael Hardy Mar 11 '14 at 22:22

2 Answers2

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By looking at upper and lower Riemann sums on this decreasing function, you can bracket it by $$\int_1^{n} x^{-\frac 12}\ dx \lt \sum_{k=1}^nk^{-\frac 12} \lt \int_0^{n-1} x^{-\frac 12}\ dx\\2\sqrt n-2\lt \sum_{k=1}^nk^{-\frac 12} \lt2\sqrt {n-1}$$

Ross Millikan
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$$ \frac \beta{k^\alpha} - \frac \beta{(k+1)^\alpha} \sim \frac {\alpha\beta}{k^{\alpha + 1}} = \frac 1{\sqrt{k}} $$if $\alpha = -1/2$ and $\beta = -2$

now as $\sum \frac 1{\sqrt{k}} = \infty$:

$$ \sum_{k=1}^N \frac 1{\sqrt{k}}\sim \sum_{k=1}^N \frac \beta{k^\alpha} - \frac \beta{(k+1)^\alpha} \sim -\frac \beta{N^\alpha} = 2\sqrt{N} $$

mookid
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