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I am studying mathematical analysis, and I wonder what is the order of the series $$ \sum_{k=1}^n\frac{1}{\sqrt{k}}. $$ I am able to show that it is $O(\sqrt{n})$: $$ \sum_{k=1}^n\frac{1}{\sqrt{k}}=2\sum_{k=1}^n\frac{1}{2\sqrt{k}}\le2\sum_{k=1}^n\frac{1}{\sqrt{k-1}+\sqrt{k}}=2\sum_{k=1}^n(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}. $$ Is the order tight? That is, is it true that $$ \sum_{k=1}^n\frac{1}{\sqrt{k}}=\Theta(\sqrt{n})? $$

Kittayo
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    of course - use the integral test to bound the sum between the integral of $1/\sqrt x$ taken from $1$ to $n-1$ and $2$ to $n$ respectively – Conrad Jan 03 '23 at 04:59
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    You are right. Alternatively, $\sum_{k=1}^n1/\sqrt{k}\ge\sum_{k=1}^n1/\sqrt{n}=\sqrt{n}$. – Kittayo Jan 03 '23 at 05:01
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    In fact, it is easy to show, for example by comparing the area under the graph of $y=\frac{1}{\sqrt{x}}$ and $y=\frac{1}{\sqrt{\lceil x\rceil}}$, that $$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}-\int_{0}^{n}\frac{1}{\sqrt{x}},\mathrm{d}x$$ is not only bounded but convergent as $n\to\infty$. (For the interested user, the limit turns out to be the value $\zeta(\frac{1}{2})$ of the Riemann zeta function.) – Sangchul Lee Jan 03 '23 at 05:19
  • According to this answer, the sum has asymptotic expansion: $$\sum_{k=1}^n \frac{1}{\sqrt{k}} \asymp 2\sqrt{n}
    • \zeta\left(\frac12\right)
    • \frac{1}{2\sqrt{n}}

    -\frac{1}{24}{n}^{-\frac32}+{\frac {1}{384}},{n}^{-\frac72} + \cdots$$

     – achille hui Jan 03 '23 at 14:28
  • @Conrad could you please tell me where "the integral test" can be found in modern textbooks? We learned precisely that, including endpoint shifts, using simple diagrams; but I can't seem to find that exact statement. The version I paste into answers is: If we have $g(x) > 0$ but $g'(x) < 0,$ then $$ \int_a^{b+1} ; g(x) ; dx ; < ; \sum_{j=a}^b ; g(j) ; < ; \int_{a-1}^b ; g(x) ; dx $$ – Will Jagy Jan 03 '23 at 18:01
  • @Will I don't know offhand where one finds an explicitly stated version but the integral test works when $g$ is integrable and monotonic, one doesn't need positivity or smoothness; if $g$ decreasing your formula above applies; if $g$ increasing the inequalities reverse since now $ \int_{k-1}^{k}g(x)dx \le g(k) \le \int_k^{k+1}g(x)dx$ - also we can adjust the endpoints as needed to get tighter inequalities if needed, while if we know more about $g$ (eg convexity) we can do even tighter bounds with half-integer intervals – Conrad Jan 03 '23 at 18:18
  • @Conrad thank you. I should have said that this was in high school, close to 50 years ago. I also like the part about using half integers. The other direction, from a text file I keep handy for Latex/Mathjax code: if we have $f(x) > 0$ and $f'(x) > 0,$ then $$ \int_{a-1}^{b} ; f(x) ; dx ; < ; \sum_{j=a}^b ; f(j) ; < ; \int_{a}^{b+1} ; f(x) ; dx $$ – Will Jagy Jan 03 '23 at 18:24

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