Factorial Series

Calculus Level 3

1 ! 4 ! + 2 ! 5 ! + 3 ! 6 ! + 4 ! 7 ! + ⋯ \large\dfrac{1!}{4!}+\dfrac{2!}{5!}+\dfrac{3!}{6!}+\dfrac{4!}{7!}+\cdots

If the value of the above series can be expressed to a b \dfrac{a}{b} when a a and b b are coprime positive integers. Find the value of a + b a+b .


The answer is 13.

Ikkyu San by Ikkyu San , 23, Thailand

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2 solutions

Tanishq Varshney
Jun 13, 2015

The given series can be written as

∑ n = 1 ∞ n ! ( n + 3 ) ! \large{\displaystyle \sum^{\infty}_{n=1} \frac{n!}{(n+3)!}}

∑ n = 1 ∞ 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) \large{\displaystyle \sum^{\infty}_{n=1} \frac{1}{(n+1)(n+2)(n+3)}}

1 2 ∑ n = 1 ∞ 1 ( n + 1 ) ( n + 2 ) − 1 ( n + 2 ) ( n + 3 ) \large{\frac{1}{2} \displaystyle \sum^{\infty}_{n=1} \frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}}

Its a telescopic series

On solving we get

1 12 \huge{\boxed{ \frac{1}{12}}}

12 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Careful there, you should put parenthesis!

Yes, telescoping series is the key to this problem.

Noel Lo
Jun 15, 2015

We get 1 2 ∗ 3 ∗ 4 + 1 3 ∗ 4 ∗ 5 + 1 4 ∗ 5 ∗ 6 + . . . \frac{1}{2*3*4} + \frac{1}{3*4*5} + \frac{1}{4*5*6} + ...

= 1 2 ∗ 2 − 1 3 + 1 2 ∗ 4 = \frac{1}{2*2} - \frac{1}{3} + \frac{1}{2*4}

+ 1 2 ∗ 3 − 1 4 + 1 2 ∗ 5 +\frac{1}{2*3} - \frac{1}{4} + \frac{1}{2*5}

+ 1 2 ∗ 4 − 1 5 + 1 2 ∗ 6 + . . . +\frac{1}{2*4} - \frac{1}{5} + \frac{1}{2*6} + ...

= 1 2 ∗ 2 − 1 3 + 1 2 ∗ 3 = 1 4 − 1 3 + 1 6 = 1 12 = \frac{1}{2*2} - \frac{1}{3} + \frac{1}{2*3} = \frac{1}{4} - \frac{1}{3}+\frac{1}{6} = \frac{1}{12}

a + b = 1 + 12 = 13 a+b = 1+12 = \boxed{13}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

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