Easy sum!

Algebra Level 3

∑ n = 1 ∞ n 3 n ! = ? \large \sum_{n=1}^\infty \dfrac{n^3}{n!} =\, ?

Give your answer to 2 decimal places.


The answer is 13.59.

Abdelghani Ssoris by Abdelghani ßoris , 24, Morocco

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2 solutions

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To your final answer it seems like comma (,) instead of 13.59

Naren Bhandari - 3 years, 10 months ago
Naren Bhandari
Jul 28, 2017

∑ n = 1 ∞ n 3 n ! = ∑ n = 1 ∞ n 2 − 1 + 1 ( n − 1 ) ! = ∑ n = 1 ∞ ( n + 1 ( n − 2 ) ! + 1 ( n − 1 ) ! ) \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)

= ∑ n = 1 ∞ ( 1 ( n − 3 ) ! + 3 ( n − 2 ) ! + 1 ( n − 1 ) ! ) = 5 e ≈ 13.59 =\displaystyle\sum_{n=1}^{\infty}\left (\frac{1}{(n-3)!} + \frac {3}{(n-2)!} + \frac {1}{(n-1)!}\right) = 5e\approx \boxed {13.59}

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