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Find the value of $$\left[\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.....+\frac{1}{\sqrt {1000}}\right]$$.Where [•] denote the greatest integer function.

I am very confused about this problem. I tried to find the upper and lower bound of the function. But I can't find any formulas to find the bounds. Somebody please help me.

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    $\begingroup$ Hint: Consider an associated integral. $\endgroup$ – quasi Mar 2 '17 at 12:36
  • $\begingroup$ That is a perfectly viable way, but then you have to prove that the difference between the sum and the integral is small. As a matter of fact, we may avoid integrals at all. $\endgroup$ – Jack D'Aurizio Mar 2 '17 at 13:02
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    $\begingroup$ Please, use $\left\lfloor\cdots\right\rfloor$ for the $\texttt{floor}$ function. For instance, $\LaTeX$ code for $\left\lfloor x\right\rfloor$ is written in $\LaTeX$ or/and $\texttt{MathJax}$ as $\texttt{\left\lfloor x\right\rfloor}$ $\endgroup$ – Felix Marin Mar 3 '17 at 1:31
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    $\begingroup$ @Felix Marin yeah an telling about the floor function. $\endgroup$ – Sufaid Saleel Mar 3 '17 at 12:23
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This is all about providing an accurate approximation for the involved generalized harmonic sum. I will use a technique (creative telescoping) clearly outlined in the first chapter of these course notes.
We may notice that $$ \sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}} = \frac{1}{\sqrt{n+\frac{1}{2}}+\sqrt{n-\frac{1}{2}}} $$ is a telescopic term and it is, additionally, pretty close to $\frac{1}{2\sqrt{n}}$. In particular $$ \frac{1}{\sqrt{n}}-2\left(\sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}}\right)= d_n\\=-\frac{1}{2\sqrt{n}\left(\sqrt{n+1/2}+\sqrt{n-1/2}\right)^2\left(\sqrt{n+1/2}+\sqrt{n}\right)\left(\sqrt{n-1/2}+\sqrt{n}\right)}$$ is a negative term that behaves like $-\frac{1}{32 n^{5/2}}$ for large $n$s. It follows that $$ \sum_{k=2}^{1000}\frac{1}{\sqrt{k}} = 2\sum_{k=2}^{1000}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right)+\sum_{k=2}^{1000}d_k $$ has a distance from $$ 2\sum_{k=2}^{1000}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right) = 2\left(\sqrt{1000+\frac{1}{2}}-\sqrt{2-\frac{1}{2}}\right) $$ that is less$^{(*)}$ than $\frac{8}{1000}$. By computing the last quantity it follows that the answer is $\color{red}{60}$.

$(*)$ Proof: we have $$ |d_k|\leq \frac{1}{48 k^{3/2}}-\frac{1}{48(k+1)^{3/2}} $$ hence $$ \sum_{k=2}^{1000}|d_k|\leq \sum_{k\geq 2}|d_k|\leq \frac{1}{48\cdot 2^{3/2}}<\frac{8}{1000}.$$

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  • $\begingroup$ Please tell what you do with this $$\sum_{k=2}^{1000}d_k $$. $\endgroup$ – Sufaid Saleel Mar 2 '17 at 15:08
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    $\begingroup$ @SufaidSaleel: all right, answer updated. $\endgroup$ – Jack D'Aurizio Mar 2 '17 at 15:15
  • $\begingroup$ @ Jack D'Aurizio check my answer I found different answer $\endgroup$ – Sufaid Saleel Mar 2 '17 at 16:34
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    $\begingroup$ @SufaidSaleel: that is no wonder since you took $10000$ instead of $1000$. $\endgroup$ – Jack D'Aurizio Mar 2 '17 at 16:46
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With the identity $\ds{\sum_{k = 1}^{N}{1 \over n^{s}} = {N^{1 - s} \over 1 - s} + \zeta\pars{s} + s\int_{N}^{\infty}{\braces{x} \over x^{s + 1}}\,\dd x}$:

\begin{align} \left\lfloor\sum_{n = 2}^{1000}{1 \over \root{n}}\right\rfloor & = \left\lfloor-1 + \sum_{n = 1}^{1000}{1 \over \root{n}}\right\rfloor = \left\lfloor-1 + \pars{2\root{1000} + \zeta\pars{1 \over 2} + {1 \over 2}\int_{1000}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x}\right\rfloor \\[5mm] & = \left\lfloor\underbrace{20\root{10} + \zeta\pars{1 \over 2} - 1} _{\ds{\approx\ \color{#f00}{60.7852}}}\ +\ \underbrace{{1 \over 2}\int_{1000}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x} _{\ds{\left\vert\begin{array}{l}\ds{> 0} \\ \mbox{and}\ <\ {\root{10} \over 100}\ \approx\ \color{#f00}{0.0316} \end{array}\right.}}\right\rfloor = \bbx{\ds{\large\color{#f00}{60}}} \end{align}

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The inequality

$${1\over\sqrt n}\gt\int_n^{n+1}{dx\over\sqrt x}=2(\sqrt{n+1}-\sqrt n)$$

is geometrically obvious, since $1\over\sqrt x$ is a decreasing function. The inequality

$${1\over\sqrt n}\lt\int_{n-{1\over2}}^{n+{1\over2}}{dx\over\sqrt x}=2\left(\sqrt{n+{1\over2}}-\sqrt{n-{1\over2}}\right)$$

is not obvious, but straightforward to verify algebraically:

$$\begin{align} {1\over\sqrt n}\lt2\left(\sqrt{n+{1\over2}}-\sqrt{n-{1\over2}}\right) &\iff{1\over\sqrt{2n}}\lt\sqrt{2n+1}-\sqrt{2n-1}\\ &\iff{1\over2n}\lt(2n+1)-2\sqrt{4n^2-1}+(2n-1)\\ &\iff\sqrt{4n^2-1}\lt2n-{1\over4n}\\ &\iff4n^2-1\lt4n^2-1+{1\over16n^2} \end{align}$$

We therefore get

$$\int_2^{1001}{dx\over\sqrt x}\lt{1\over\sqrt2}+{1\over\sqrt3}+\cdots+{1\over\sqrt{1000}}\lt\int_{3/2}^{2001/2}{dx\over\sqrt x}$$

From $\int_2^{1001}{dx\over\sqrt x}=2(\sqrt{1001}-\sqrt2)\approx60.4487$ and $\int_{3/2}^{2001/2}{dx\over\sqrt x}=2(\sqrt{2001/2}-\sqrt{3/2})\approx60.81187$ we find

$$\left\lfloor{1\over\sqrt2}+{1\over\sqrt3}+\cdots+{1\over\sqrt{1000}}\right\rfloor=60$$

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  • $\begingroup$ Confirmed with Mathematica: N@Sum[1/Sqrt[j], {j, 2, 1000}] = 60.801. $\endgroup$ – David G. Stork May 1 '17 at 23:44
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$$\sum_{k=2}^{1000}\frac1{\sqrt k}=\sum_{a=2}^{10}\frac1{\sqrt a}+\sum_{b=11}^{1000}\frac1{\sqrt b}$$

$$\sum_{a=2}^{10}\frac1{\sqrt a}\approx4.02100$$

Now notice that

$$\sqrt k-\sqrt{k-1}<\frac1{\sqrt k}<\sqrt{k+1}-\sqrt k$$

Which gives a telescoping series and...

$$56.64392\approx2(\sqrt{1001}-\sqrt{11})<\sum_{b=11}^{1000}\frac1{\sqrt b}<2(\sqrt{1000}-\sqrt{10})\approx56.92100$$

Thus,

$$60.66492<\sum_{k=2}^{1000}\frac1{\sqrt k}<60.94200$$

And so we see the solution is $60$.

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