Simplify the ratio

Oct 2014
57
0
Nashik
1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ...... + 1/(100*101*102)

thank you
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey Nashik2014.

Results like these typically use identities of some sort.

You could however look at starting with two terms and looking for an inductive argument.

Do you know how to construct and test induction based results?
 

Jester

MHF Helper
Dec 2008
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1,255
Conway AR
Are you familiar with sigma notation. For example, your sum can be written as

\(\displaystyle \sum_{n=1}^{100} \dfrac{1}{n(n+1)(n+2)}\)
 

skeeter

MHF Helper
Jun 2008
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6,765
North Texas
using partial fractions ...

$\displaystyle \sum_{n=1}^{100} \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \sum_{n=1}^{100} \frac{1}{n} - \sum_{n=1}^{100} \frac{1}{n+1} + \frac{1}{2} \sum_{n=1}^{100} \frac{1}{n+2}$

$\displaystyle \frac{1}{2}\bigg[1+\frac{1}{2}+\frac{1}{3}+ \, ... \, + \frac{1}{99} + \frac{1}{100}\bigg] - \bigg[\frac{1}{2}+\frac{1}{3}+ \frac{1}{4} \, ... \, + \frac{1}{100} + \frac{1}{101}\bigg] + \frac{1}{2}\bigg[\frac{1}{3}+\frac{1}{4}+ \, ... \, + \frac{1}{101} + \frac{1}{102}\bigg]$

multiply each summation by 2 ...

$\displaystyle \bigg[1+\frac{1}{2}+\frac{1}{3}+ \, ... \, + \frac{1}{99} + \frac{1}{100}\bigg] - \bigg[\frac{2}{2}+\frac{2}{3}+ \frac{2}{4} \, ... \, + \frac{2}{100} + \frac{2}{101}\bigg] + \bigg[\frac{1}{3}+\frac{1}{4}+ \, ... \, + \frac{1}{101} + \frac{1}{102}\bigg]$

after a whole lot of zero cancellations, the sum of the above is $\bigg[1-\dfrac{1}{2}-\dfrac{1}{101}+\dfrac{1}{102}\bigg]$ ... all that's left is to divide this result by 2
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
using partial fractions ...

$\displaystyle \sum_{n=1}^{100} \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \sum_{n=1}^{100} \frac{1}{n} - \sum_{n=1}^{100} \frac{1}{n+1} + \frac{1}{2} \sum_{n=1}^{100} \frac{1}{n+2}$

$\displaystyle \frac{1}{2}\bigg[1+\frac{1}{2}+\frac{1}{3}+ \, ... \, + \frac{1}{99} + \frac{1}{100}\bigg] - \bigg[\frac{1}{2}+\frac{1}{3}+ \frac{1}{4} \, ... \, + \frac{1}{100} + \frac{1}{101}\bigg] + \frac{1}{2}\bigg[\frac{1}{3}+\frac{1}{4}+ \, ... \, + \frac{1}{101} + \frac{1}{102}\bigg]$

multiply each summation by 2 ...

$\displaystyle \bigg[1+\frac{1}{2}+\frac{1}{3}+ \, ... \, + \frac{1}{99} + \frac{1}{100}\bigg] - \bigg[\frac{2}{2}+\frac{2}{3}+ \frac{2}{4} \, ... \, + \frac{2}{100} + \frac{2}{101}\bigg] + \bigg[\frac{1}{3}+\frac{1}{4}+ \, ... \, + \frac{1}{101} + \frac{1}{102}\bigg]$

after a whole lot of zero cancellations, the sum of the above is $\bigg[1-\dfrac{1}{2}-\dfrac{1}{101}+\dfrac{1}{102}\bigg]$ ... all that's left is to divide this result by 2
Kinda nice the way that worked out, huh!
 
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skeeter

MHF Helper
Jun 2008
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North Texas
Kinda nice the way that worked out, huh!
Don't recall harmonic series having a canned formula, so I remember what my Calc 2 prof said a long time ago ... sometimes you have to list out the summation(s) make stuff happen.