Sum of the Given Series (7)

Calculus Level 3

1 3 1 ! + 2 3 2 ! + 3 3 3 ! + ⋯ = ? \large\frac {1^3}{1!}+\frac {2^3}{2!}+\frac {3^3}{3!}+\cdots=\, ?

Try this problem first: Sum of the Given Series (6) .


The answer is 13.5914.

Brian Lie by Brian Lie , 26, China

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1 solution

Naren Bhandari
May 23, 2018

∑ n = 1 ∞ n 3 n ! = ∑ n = 1 ∞ n 2 − 1 + 1 ( n − 1 ) ! = ∑ n = 1 ∞ ( n + 1 ( n − 2 ) ! + 1 ( n − 1 ) ! ) = ∑ n = 1 ∞ ( 1 ( n − 3 ) ! + 2 ( n − 2 ) ! + 1 ( n − 2 ) ! + 1 ( n − 1 ) ! ) = 5 e ≈ 13.5914 \displaystyle\sum_{n=1}^{\infty}\frac {n^3}{n!} = \displaystyle\sum_{n=1}^{\infty}\frac {n^2-1+1}{(n-1)!} =\displaystyle\sum_{n=1}^{\infty}\left (\frac {n+1}{(n-2)!} + \frac {1}{(n-1)!}\right) \\ =\displaystyle\sum_{n=1}^{\infty}\left (\frac{1}{(n-3)!} + \frac{2}{(n-2)!}+ \frac {1}{(n-2)!} + \frac {1}{(n-1)!}\right) = 5e\approx \boxed {13.5914}

Alternative approach is using Bell number .

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