Difference between revisions of "2018 MPFG Problem 19"
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We first applicate the right Riemann sum of <math>y=\frac{1}{\sqrt{x}}</math> | We first applicate the right Riemann sum of <math>y=\frac{1}{\sqrt{x}}</math> | ||
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<cmath>2S_n > \int{f_{1}^{9803}(\frac{1}{\sqrt{x}})}dx</cmath> | <cmath>2S_n > \int{f_{1}^{9803}(\frac{1}{\sqrt{x}})}dx</cmath> | ||
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Then applicate the left Riemann sum of <math>y=\frac{1}{\sqrt{x}}</math> | Then applicate the left Riemann sum of <math>y=\frac{1}{\sqrt{x}}</math> | ||
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<cmath>2S_n-1 < \int{f_{1}^{9801}(\frac{1}{\sqrt{x}})}dx</cmath> | <cmath>2S_n-1 < \int{f_{1}^{9801}(\frac{1}{\sqrt{x}})}dx</cmath> | ||
Revision as of 06:15, 24 August 2025
Problem 19
Consider the sum
![\[S_n = \sum_{k=1}^{n}\frac{1}{\sqrt{2k-1}}\]](http://latex.artofproblemsolving.com/2/4/0/240b62e32ea68da7b45be64107ac9da7c52d1b8e.png)
Determine 
. Recall that if 
is a real number, then 
(the floor of x) is the greatest integer that is less than or equal to .
Solution 1
We can think of this problem through integration perspectives. Observe that 
looks very similar to a Riemann sum.
![\[_n = 1/\sqrt{1}+1/\sqrt{3}+ ... + 1/\sqrt{9801}\]](http://latex.artofproblemsolving.com/6/5/1/651fe8f1d84134e18573f616c24d72d631563e0d.png)
We first applicate the right Riemann sum of 
![\[2S_n > \int{f_{1}^{9803}(\frac{1}{\sqrt{x}})}dx\]](http://latex.artofproblemsolving.com/8/0/1/8017cbcc2cbdc76d70bd964a89a88c6339f11c55.png)
![\[2S_n > \left. (2x^{\frac{1}{2}})\right|_{1}^{9803}\]](http://latex.artofproblemsolving.com/3/f/2/3f239abd737ba3da55d27fd1af4afc2ea033ff71.png)
![\[2S_n > 2(\sqrt{9803}-1)\]](http://latex.artofproblemsolving.com/d/5/b/d5b9cabd1b9220136c997c436b0f9c1e9fcf8ff0.png)
![\[S_n > \sqrt{9803}-1\]](http://latex.artofproblemsolving.com/c/c/8/cc8a19de16c79457398d7200f4e7d3372111a206.png)
Then applicate the left Riemann sum of
![\[2S_n-1 < \int{f_{1}^{9801}(\frac{1}{\sqrt{x}})}dx\]](http://latex.artofproblemsolving.com/c/6/d/c6d0f3b0e677476ee0b29caf8df29e1b57a01e46.png)
![\[2S_n-1 < \left. (2x^{\frac{1}{2}})\right|_{1}^{9801}\]](http://latex.artofproblemsolving.com/4/1/3/41370ace1d9c0be85df031a0c95411d63b92bc67.png)
![\[S_n-\frac{1}{2} > \sqrt{9801}-1\]](http://latex.artofproblemsolving.com/8/1/c/81c52b56cae20dac843c916a2ceaf4261c215970.png)
![\[S_n < \sqrt{9801}-1/2\]](http://latex.artofproblemsolving.com/7/c/e/7cee97b253a049593b963f418beedaffd44bfece.png)
We conclude that:
~cassphe
