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I need to find the partial sum formula for $\sum\limits_{n=0}^\infty \frac{n^3}{n!}$

I started by calculating some elements of the formula, but I could not find any possible patterns.

Could you please help me in finding the partial sum formula? I need it in order to be able to calculate the sum of the series. If there is any easier way to do this than finding a partial sum formula, please let me know.

Thank you in advance

user
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coki1405
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3 Answers3

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We have that

$$\sum_{n=0}^\infty \frac{n^3}{n!}=\sum_{n=1}^\infty \frac{n^2}{(n-1)!}= \sum_{n=1}^\infty \frac{n^2-n+n-1}{(n-1)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=$$

$$=\sum_{n=2}^\infty \frac{n-2+2}{(n-2)!}+\sum_{n=2}^\infty \frac{1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=$$

$$=\sum_{n=3}^\infty \frac{1}{(n-3)!}+3\sum_{n=2}^\infty \frac{1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=5e$$

user
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    Cute trick! Keeping it elementary. – Aryabhata Nov 02 '19 at 00:44
  • @Aryabhata Thanks! Also your solution is a valuable way! – user Nov 02 '19 at 00:47
  • Thank you! But I do not quite get what you do from line 1 to line 2. where does the second term ∑1/(n−2)! come from? And in the third line, why is there a 3 in front of the second sum? – coki1405 Nov 02 '19 at 09:12
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    We have $$\sum_{n=1}^\infty \frac{n^2-n+n-1}{(n-1)!}=\sum_{n=1}^\infty \frac{n^2-n}{(n-1)!}+\sum_{n=1}^\infty \frac{n-1}{(n-1)!}=\sum_{n=2}^\infty \frac{n}{(n-2)!}+\sum_{n=2}^\infty \frac{1}{(n-2)!}$$ – user Nov 02 '19 at 09:14
  • Okay, I understand. and the 3 in the last line? – coki1405 Nov 02 '19 at 09:25
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    $$\sum_{n=2}^\infty \frac{n-2+2}{(n-2)!}=\sum_{n=2}^\infty \frac{n-2}{(n-2)!}+\sum_{n=2}^\infty \frac{2}{(n-2)!}=\sum_{n=3}^\infty \frac{1}{(n-3)!}+2\sum_{n=2}^\infty \frac{1}{(n-2)!}$$ – user Nov 02 '19 at 09:29
  • Thank you so much!! – coki1405 Nov 02 '19 at 09:32
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Start with $$xe^{x} = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}$$

Differentiate, multiple by $x$ differentiate again and set $x = 1$.

I believe the answer comes out to $5e$.

Aryabhata
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Just another way to look at it.

Considering $$\sum_{n=0}^\infty \frac {n^3} {n!} x^n$$ write first $$n^3=n(n-1)(n-2)+3n(n-1)+n$$ $$\sum_{n=0}^\infty \frac {n^3} {n!} x^n=x^3\sum_{n=0}^\infty \frac {n(n-1)(n-2)} {n!} x^{n-3}+3x^2\sum_{n=0}^\infty \frac {n(n-1)} {n!} x^{n-2}+x\sum_{n=0}^\infty \frac {n} {n!} x^{n-1}$$ that is to say $$\sum_{n=0}^\infty \frac {n^3} {n!} x^n=x^3\left(\sum_{n=0}^\infty \frac {x^n} {n!} \right)'''+3x^2\left(\sum_{n=0}^\infty \frac {x^n} {n!} \right)''+x\left(\sum_{n=0}^\infty \frac {x^n} {n!} \right)'=(x^3+3x^2+x)\,e^x$$

Now, make $x=1$ to get the result.

Claude Leibovici
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