Algebra #22

by sjaelee, Sep 24, 2011, 3:11 PM

Quote:
Determine
\[ \left\lfloor 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{1000000}}\right\rfloor. \]

We start by making bounds. We note that

\[\sum_{k=1}^{n}\frac{1}{\sqrt{k}+\sqrt{k+1}}\]

telescopes as \[\sqrt{k+1}-\sqrt{k}\]

and is very close to our original sum divided by two, slighly less.

Evaulauting, we have t (with $n=1000000$)

\[\sqrt{1000001}-1\]. Multiplying 2, we have a number $x$ such that

$198<x<199$. We know our desired is greater.

Now, the lower bound. I once proved that

\[ 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n} \]

Proof

We can prove for postive integers $n\ge2$ that

\[ 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}-1 \]

in the same way if we find a base case. Testing, $2$ works.

Note that $2\sqrt{1000000}-1=199$. Thus, our desired, $y$,

$198<y<199$. Thus, our greatest integer is

$198$.
This post has been edited 4 times. Last edited by djmathman, Sep 24, 2011, 4:17 PM

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  • legendary problem writer

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  • hello :)

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  • Do you have a link to your main blog that you started after graduating from high school, I couldn't find it. @dj I met you IRL at Awesome Math summer Program several years ago.

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