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I am looking for an upper bound of $\sum_{k=1}^n \frac{1}{\sqrt{k}}$. Alternatively, is the sequence $\frac{1}{n\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt{k}}$ bounded?

I am trying to use a Strong law of Large Numbers by Feller and need to show this condition.

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  • $\begingroup$ Maybe specify a bit about what you are looking for would help. Say, $k$ is a straightforward upper bound for the sum... $\endgroup$
    – astro
    Jul 4, 2020 at 18:48
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    $\begingroup$ Note: $$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\leq\int_{0}^{n}\frac{\mathrm{d}x}{\sqrt{x}}=2\sqrt{n}.$$ $\endgroup$ Jul 4, 2020 at 18:49
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    $\begingroup$ Not in terms of $O(1)$ but the bound is of form $\sum_{n=1}^{k}\frac{1}{\sqrt{n}}=O(\sqrt{k})$. $\endgroup$
    – mertunsal
    Jul 4, 2020 at 18:50
  • $\begingroup$ Thanks! That's what I needed. $\endgroup$
    – adalaio
    Jul 4, 2020 at 18:55
  • $\begingroup$ I meant $n{}{}$ $\endgroup$
    – astro
    Jul 5, 2020 at 1:00

4 Answers 4

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Via a comparison series and integral: $$ \sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \int_{k-1}^{k}\frac{dx}{\sqrt{k}} \leq \sum_{k=1}^n \int_{k-1}^{k}\frac{dx}{\sqrt{x}} = \int_{0}^{n}\frac{dx}{\sqrt{x}} = 2\sqrt{n} $$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets use a powerful Riemann Zeta Function Identity:

\begin{align} \sum_{k = 1}^{n}{1 \over \root{k}} & = \sum_{k = 1}^{n}{1 \over k^{\color{red}{1/2}}} = {n^{1 - \color{red}{1/2}} \over 1 - \color{red}{1/2}} + \zeta\pars{\color{red}{1 \over 2}} + \color{red}{1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{\color{red}{1 /2} +1}}\,\dd x \\[5mm] & = 2\root{n} + \zeta\pars{1 \over 2} + {1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x \end{align} However, $$ {1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x < {1 \over 2}\int_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}} $$


$$ \\ \mbox{Then,}\quad \bbx{\sum_{k = 1}^{n}{1 \over \root{k}} < 2\root{n} + \zeta\pars{1 \over 2} + {1 \over \root{n}}}\\ $$ Note that $\ds{\zeta\pars{1/2} < 0}$.
$$ \mbox{Also,}\quad \sum_{k = 1}^{n - 1}{1 \over \root{k}} + {1 \over \root{n}} < 2\root{n} + \zeta\pars{1 \over 2} + {1 \over \root{n}} $$ $$ \\ \mbox{which leads to}\quad \bbx{\sum_{k = 1}^{n}{1 \over \root{k}} < 2\root{n + 1} + \zeta\pars{1 \over 2}}\\ $$
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$\sum_{k=1}^n \frac{1}{\sqrt{k}}$ is a $p$-series with $p\le 1$, hence it diverges.

Alternatively, $$0<\frac{1}{n\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt 1}=\frac1{\sqrt n}\le1.$$

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  • $\begingroup$ Thanks! Indeed, I know it diverges. I needed an upper bound on the rate at which it diverges, and learned that an upper bound on the rate is $\sqrt{n}$. I am a bit confused by your proof of the upper bound of the sequence. Can you please explain? $\endgroup$
    – adalaio
    Jul 4, 2020 at 19:44
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Hint:

For $k\le x<k+1$,

$$\frac1{\sqrt k}<\frac1{\sqrt{x-1}}.$$

Integrate from $1$ to $n+1$.

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