Factorial Sum II

Calculus Level 4

∑ n = 1 ∞ ( n ) ( n + 1 ) ( 2 n + 1 ) 6 ( n ! ) = A e B \sum_{n=1}^{\infty}\frac{(n)(n+1)(2n+1)}{6(n!)}=\frac{Ae}{B} where A , B A,B are positive co-prime integers and e e is Euler constant. Then find A + B A+B .


The answer is 23.

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Shivam Jadhav
Aug 17, 2015

∑ k = 0 ∞ n ( n + 1 ) ( 2 n + 1 ) 6 n ! 1 6 ∑ k = 0 ∞ 2 n 3 + 3 n 2 + n n ! 1 3 ∑ k = 0 ∞ n 3 n ! + 1 2 ∑ k = 0 ∞ n 2 n ! + 1 6 ∑ k = 0 ∞ n n ! e ( 5 3 + 2 2 + 1 6 ) 17 6 e \sum_{k = 0}^{\infty}{\dfrac{n(n+1)(2n+1)}{6 n!}} \\\frac{1}{6}\sum_{k = 0}^{\infty}{\dfrac{2n^3 + 3n^2 + n}{n!}} \\ \frac{1}{3}\sum_{k=0}^{\infty}{\dfrac{n^3}{n!}} + \frac{1}{2}\sum_{k=0}^{\infty}{\dfrac{n^2}{n!}} + \frac{1}{6}\sum_{k=0}^{\infty}{\dfrac{n}{n!}} \\ e \left(\frac{5}{3} + \frac{2}{2} + \frac{1}{6} \right) \\ \boxed{\dfrac{17}{6}e}

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