Factorial Sum!

Algebra Level 3

1 2 ! + 2 3 ! + 3 4 ! + … + 99 100 ! = ? \large \dfrac1{2!}+\dfrac2{3!}+\dfrac3{4!}+\ldots+\dfrac{99}{100!} = \ ?

100 ! 100! Cannot be determined 1 1 1 − 1 100 ! 1-\frac1{100!} 1 + 1 100 ! 1+\frac1{100!}

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Naitik Sanghavi by naitik sanghavi , 20, India

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4 solutions

We have:

1 2 ! + 2 3 ! + 3 4 ! + … + 99 100 ! \quad\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\ldots+\dfrac{99}{100!}

= 2 − 1 2 ! + 3 − 1 3 ! + 4 − 1 4 ! + … + 100 − 1 100 ! =\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+\ldots+\dfrac{100-1}{100!}

= 1 1 ! − 1 2 ! + 1 2 ! − 1 3 ! + 1 3 ! − 1 4 ! + … + 1 99 ! − 1 100 ! =\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+\ldots+\dfrac{1}{99!}-\dfrac{1}{100!}

= 1 − 1 100 ! =1-\dfrac{1}{100!} .

34 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Yes. This is just a telescoping sum hiding in plain sight.

I tried placing this in a excel formula but Excel has limitation and will not allow me to calculate 29/30! or further... but summing up the first 28 gives me an answer of 1... for anyone who jumped to excel to do a simple formula with rows of numbers (as I just did) will be falsely given an answer of 1 ... and Excel is wrong... the answer is 1-(1/(100!))... when I put 1-(1/(100!)) in to excel I get the answer of 1... but anyone can obviously see that 1 minus something will never give you 1 unless that something is 0. and the only way 1 over something will give you zero is if that something is zero... and everybody knows 100! is not zero... Excel has limitations and I am vastly disappointed in knowing how screwed up Excel is in math computations. But then again computers are all 1's and 0's... how can we get them to compute an analogue stream of numbers?

Jason Proffitt - 5 years, 10 months ago
Kenny Lau
Aug 11, 2015

1 2 ! + 2 3 ! + 3 4 ! + … + 99 100 ! \quad\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\ldots+\dfrac{99}{100!}

= ∑ n = 1 99 n ( n + 1 ) ! =\displaystyle\sum_{n=1}^{99} \frac{n}{(n+1)!}

= ∑ n = 1 99 ( n + 1 ( n + 1 ) ! − 1 ( n + 1 ) ! ) =\displaystyle\sum_{n=1}^{99} \left( \frac{n+1}{(n+1)!} - \frac{1}{(n+1)!} \right)

= ∑ n = 1 99 ( 1 n ! − 1 ( n + 1 ) ! ) =\displaystyle\sum_{n=1}^{99} \left( \frac{1}{n!} - \frac{1}{(n+1)!} \right)

= ∑ n = 1 99 1 n ! − ∑ n = 1 99 1 ( n + 1 ) ! =\displaystyle\sum_{n=1}^{99} \frac{1}{n!} - \sum_{n=1}^{99} \frac{1}{(n+1)!}

= ∑ n = 1 99 1 n ! − ∑ n = 2 100 1 n ! =\displaystyle\sum_{n=1}^{99} \frac{1}{n!} - \sum_{n=2}^{100} \frac{1}{n!}

= ∑ n = 1 1 1 n ! − ∑ n = 100 100 1 n ! =\displaystyle\sum_{n=1}^{1} \frac{1}{n!} - \sum_{n=100}^{100} \frac{1}{n!}

= 1 1 ! − 1 100 ! =\displaystyle \frac{1}{1!} - \frac{1}{100!}

= 1 − 1 100 ! =\displaystyle 1 - \frac{1}{100!}

3 Helpful 0 Interesting 1 Brilliant 0 Confused
Alan Yan
Aug 10, 2015

An Alternate Solution: After observing the pattern, We will use induction to prove the general case.

For all n ≥ 2 n \geq 2 , P ( n ) : 1 2 ! + 2 3 ! + . . . + n − 1 n ! = 1 − 1 n ! P(n): \frac{1}{2!}+\frac{2}{3!}+ ... +\frac{n-1}{n!} = 1-\frac{1}{n!}

Base Case: P ( 2 ) : 1 2 ! = 1 2 = 1 − 1 2 ! P(2): \frac{1}{2!} = \frac{1}{2} = 1-\frac{1}{2!} .

Assume P ( n ) P(n) is true. We will prove P ( n + 1 ) P(n+1) is true.

P ( n + 1 ) : 1 2 ! + 2 3 ! + . . . + n − 1 n ! + n ( n + 1 ) ! = 1 − 1 n ! + n ( n + 1 ) ! = 1 − 1 ( n + 1 ) ! P(n+1): \frac{1}{2!}+\frac{2}{3!}+...+\frac{n-1}{n!}+\frac{n}{(n+1)!} = 1-\frac{1}{n!}+\frac{n}{(n+1)!} = 1-\frac{1}{(n+1)!} Therefore by Induction, the claim is true.

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Hadia Qadir
Aug 12, 2015

he summation can be written as , 1/1! - 1/2! + 1/2! - 1/3! +1/3! - 1/4! + .... .... + 1/99! -1/100!, which ultimately gets reduced to 1 - 1/100!.

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