how to prove that$\sum_{n=1}^{\infty}{ n^4\over n!}=15e$

Question

How to prove that

$$\sum_{n=1}^{\infty}{ n^4\over n!}=15e$$

I think this is a problem of exponential series.

And now we must ask: can our over-eager answers wait until Debo has a chance to consider Daniel's hint. — GEdgar, Aug 29 '13 at 11:50

score 11 · Answer 1 · answered Aug 29 '13 at 11:50

11

Hint:

Let $$f(x)= \sum_{n=0}^\infty \frac{x^n}{n!}$$ If you differentiate it you get $$f'(x)=\sum_{n=0}^\infty n\cdot \frac{x^{n-1}}{n!}$$

Now $$x\cdot f'(x)= \sum_{n=1}^\infty n\cdot \frac{x^n}{n!}$$ If you do this several times and look at $x=1$ at the value you get your sum. (as $f(x)=\exp(x)$ you know the value of the left hand side)

answered Aug 29 '13 at 11:50

Dominic Michaelis

19,935

How do I know the values of f'(x) ,f"(x) ? – DebojyotiMukh Aug 29 '13 at 11:54
1

@DebojyotiMukh the function is the exponentialfunction and we know that $f'(x)=f(x)$. And $\exp(1)=e$ – Dominic Michaelis Aug 29 '13 at 11:57

how to prove that$\sum_{n=1}^{\infty}{ n^4\over n!}=15e$

1 Answers1