# Simplify the ratio

#### Nashik2014

1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ...... + 1/(100*101*102)

thank you

#### chiro

MHF Helper
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?

#### Nashik2014

No, give me a clue (example) please.

#### Jester

MHF Helper
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
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
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!

1 person

#### skeeter

MHF Helper
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.