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prove: 19 < S < 20

Albert

Well-known member
Jan 25, 2013
1,225
$ \text{S}=\dfrac {2008}{1000} + \dfrac {2008}{1001} + \dfrac {2008}{1002}+---------+\dfrac {2008}

{1009}$

$\text {prove} :\,\, 19 < \text {S} <20 $

note :no use of calculator or computer
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,900
Re: prove : 19< S < 20

$ \text{S}=\dfrac {2008}{1000} + \dfrac {2008}{1001} + \dfrac {2008}{1002}+---------+\dfrac {2008}

{1009}$

$\text {prove} :\,\, 19 < \text {S} <20 $

note :no use of calculator or computer
Without a calculator... but it still takes some calculations.

Since $1 - x \le \frac 1 {1+x} \le 1 - x + x^2$,

and $S = \sum_k \frac {2008}{1000 + k} = \sum_k \frac {2008}{1000} \frac {1}{1 + (k/1000)}$,

it follows that

$$\sum_k \frac {2008}{1000} (1 - \frac k {1000}) \le S \le \sum_k \frac {2008}{1000} (1 - \frac k {1000} + (\frac k {1000})^2)$$

$$\frac {2008}{1000}(10 - \frac {\sum k} {1000}) \le S \le \frac {2008}{1000}(10 - \frac {\sum k} {1000} + \frac {\sum k^2} {10^6})$$

$$\frac {2008}{1000}(10 - \frac {10\cdot (0+9)/2} {1000}) \le S \le \frac {2008}{1000}(10 - \frac {10\cdot (0+9)/2} {1000} + \frac {\frac 1 6 10(10+1)(2\cdot 10 +1)} {10^6})$$

$$20.08 - \frac {45 \cdot 2008}{10^6} \le S \le 20.08 - \frac {45 \cdot 2008}{10^6} + \frac {2008 \cdot \frac 1 6 \cdot 10 \cdot 11 \cdot 19}{10^9}$$

$$19 \le 19.98 \le S \le 19.99 + 0.0006024 \le 19.9906024 \le 20 \qquad \blacksquare$$
 
Last edited:

Albert

Well-known member
Jan 25, 2013
1,225
Re: prove : 19< S < 20

$ 19=\dfrac {10 \times 1919}{1010} < \dfrac {10 \times 2008}{1009}<S ----------(1) $

$ \therefore 19< S $

To prove S<20 is a little more tricky ,please try it ,I will upload later
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,900
Re: prove : 19< S < 20

$ 19=\dfrac {10 \times 1919}{1010} < \dfrac {10 \times 2008}{1009}<S ----------(1) $

$ \therefore 19< S $

To prove S<20 is a little more tricky ,please try it ,i will upload later
Yeah, I got that one.
It was the 20 that I found trickier.
When I got that one, I took the lower bound along, since that was part of the upper bound anyway.
 

Albert

Well-known member
Jan 25, 2013
1,225
Re: prove : 19< S < 20

$\text {Let} \,\, x=1000$
$S=(2x+8)[\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+------------+\dfrac{1}{x+9}]$
$=(2x+8)[(\dfrac{1}{x}+\dfrac{1}{x+9})+------------+(\dfrac{1}{x+4}+\dfrac{1}{x+5}) ] $
$=(2x+8)[(\dfrac{2x+9}{x^2+9x}+------------+\dfrac{2x+9}{x^2+9x+20}) ] $
$=(2x+8)(2x+9)[(\dfrac{1}{x^2+9x}+------------+\dfrac{1}{x^2+9x+20}) ] $
$<(\dfrac{4x^2+34x+72}{x^2+9x})\times 5<(\dfrac{4x^2+36x}{x^2+9x})\times 5=4\times 5=20 $
$\therefore S<20$