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Is there an easy way to calculate $f(x)=\left\lfloor{H_{x}^\left({\frac12}\right)}\right\rfloor$ for large $x\in\mathbb{N}$ where $${H_{x}^\left({\frac12}\right)}=\sum_{n=1}^xn^{-\frac12}$$ without using a calculator? For example, how to calculate $f\left({10^6}\right)$? It is known that derivative of ${H_{x}^\left({\frac12}\right)}$ is monotonically decreasing for all real positive numbers and $f(x)<2000$ for all $0<x<10^6$. I computed $f\left({10^6}\right)$ using Mathematica and it is equal to $1998$, but is it possible to find it without using programs or calculators?

2 Answers2

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According to this answer, the sum has an asymptotic expansion of the form $$\sum_{n=1}^x \frac{1}{\sqrt{n}} \approx 2\sqrt{x} + \zeta(1/2) + \frac{1}{2\sqrt{x}} -\frac{1}{24}{x}^{-3/2}+{\frac {1}{384}}\,{x}^{-7/2} + \cdots$$ If you want a quick answer for large $x$, you just keep the first two term in the expansion and get $$f\left(10^6\right) \approx \left\lfloor 2\sqrt{10^6} -1.460354508809587 \right\rfloor = 1998$$

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achille hui
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  • Can we prove that $-2<\zeta\left({\frac12}\right)<-1$ without using a calculator? –  Jan 03 '15 at 22:18
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    @Mathematician171 If you just want a bound, you can compare the sum to corresponding integral. Since $\frac{1}{\sqrt{n}}$ is monotonic decreasing, we have

    $$\sum_{n=1}^{x-1}\frac{1}{\sqrt{n}} > \int_1^x \frac{dt}{\sqrt{t}} >\sum_{n=2}^x \frac{1}{\sqrt{n}} \implies 2\sqrt{x} - 2 + \frac{1}{\sqrt{n}} < \sum_{n=1}^x \frac{1}{\sqrt{n}} < 2\sqrt{x}-1 $$

     – achille hui Jan 03 '15 at 22:34
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ If we use the Abel-Plana Formula and neglects the contribution from the integral $\ds{\pars{~\mbox{the one which involves the factor}\ \bracks{\expo{2\pi z} - 1}^{-1}~}}$ we find: \begin{align}&\color{#66f}{\large\sum_{n\ =\ 1}^{x}{1 \over \root{n}}} =\sum_{n\ =\ 0}^{\infty}\pars{{1 \over \root{n + 1}} - {1 \over \root{n + x + 1}}} \\[5mm]&\approx\int_{0}^{\infty} \pars{{1 \over \root{z + 1}} - {1 \over \root{z + x + 1}}}\,\dd z + \half\left.\pars{{1 \over \root{n + 1}} - {1 \over \root{n + x + 1}}} \right\vert_{n\ =\ 0} \\[5mm]&=\color{#66f}{\large2\pars{\root{x + 1} - 1} +\half\pars{1 - {1 \over \root{x + 1}}}} \end{align}

With $\ds{\color{#66f}{\large x = 10^{6}}}$ we got: $$ \color{#66f}{\large\sum_{n\ =\ 1}^{x}{1 \over \root{n}}} \approx {4000003 \over 2\root{1000001}} - {3 \over 2} \approx \color{#66f}{\large 1998.5005} $$
Felix Marin
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