Summing reciprocal factorials

Algebra Level 3

Find the closed form of the sum $\displaystyle \sum_{r=1}^{n} \dfrac{r}{(r+1)!}$ .

Notation : $!$ denotes the factorial function. For example, $8! = 1\times2\times3\times\cdots\times8$ .

2 - \frac1{n!}

1 - \frac2{(n+2)!}

2 - \frac1{(n+1)!}

1 - \frac1{(n+1)!}

2 solutions

Chew-Seong Cheong
May 14, 2016

$\begin{aligned} S & = \sum_{r=1}^n \frac{r}{(r+1)!} \\ & = \sum_{r=1}^n \frac{r}{r!(r+1)} \\ & = \sum_{r=1}^n \frac{1}{r!}\left(1-\frac{1}{r+1} \right) \\ & = \sum_{r=1}^n \left( \frac{1}{r!} -\frac{1}{(r+1)!} \right) \\ & = \sum_{r=1}^n \frac{1}{r!} - \sum_{r=2}^{n+1} \frac{1}{r!} \\ & = \boxed{1 - \dfrac{1}{(n+1)!}} \end{aligned}$

Ashraful Mahin
May 13, 2016

Induction can be used to solve this problem.

see S1=1/(1+1)! = 1/2 =(2-1)/2 = (2!-1)/2! ,S2=S1+2/(2+1)! =1/2 + 1/3 = 5/6 = (6-1)/6 = (3!-1)/3!..................so we can geuss that ,Sn = ((n+1)!-1)/(n+1)!

so,our inductive hepothesis is 1/(1+1)! + 2/(2+1)! +......+ n/(n+1)! = ((n+1!)-1)/(n+1)!.Now I add (n+1)/(n+1+1)! or (n+1)/(n+2)! .

so, 1/(1+1)! + 2/(2+1)! +......+n/(n+1)! + (n+1)/(n+2)!=((n+1)!-1)/(n+1)!+(n+1)/(n+2)!= ((n+1+1)! -1)/(n+1+1)! = ((n+2)!-1)/(n+2)!

the base case is S1........................so,by induction the sum is ((n+1)!-1)/(n+1)! or 1 - 1/(n+1)!

Yep, we can use induction but we need to know the answer beforehand for doing so.

Aditya Sky - 5 years, 1 month ago

Summing reciprocal factorials

2 solutions

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