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# Limits unbounded?

Find $$\sum_{n=1}^{\infty} \frac{n^2}{n!}$$.

Note by Paramjit Singh
3 years, 6 months ago

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$\begin{eqnarray} \displaystyle \sum_{n=1}^\infty \frac {n^2}{n!} & = & \sum_{n=1}^\infty \frac {n^2 - n + n}{n!} \\ & = & \sum_{n=1}^\infty \frac {n(n-1)}{n!} + \sum_{n=1}^\infty \frac {n}{n!} \\ & = & \sum_{n=2}^\infty \frac {n-1}{(n-1)!} + \sum_{n=1}^\infty \frac {1}{(n-1)!} \\ & = & \sum_{n=2}^\infty \frac {1}{(n-2)!} + \sum_{n=1}^\infty \frac {1}{(n-1)!}, \text{ let } p=n-2, \text{ let } q=n-1 \\ & = & \sum_{p=0}^\infty \frac {1}{p!} + \sum_{q=0}^\infty \frac {1}{q!} \\ & = & e +e = \boxed{2e} \\ \end{eqnarray}$

The trick to evaluate $$\displaystyle \sum_{n=1}^\infty \frac {n^k}{n!}$$ for a positive integer $$k$$ is to determine a linear combination of $$n^k$$ in terms of $$n , \space n(n-1), \space n(n-1)(n-2) , \space \ldots \space , \space n(n-1)(n-2) \cdot \cdot \cdot (n-k+1)$$

For example $$n^4 = n(n-1)(n-2)(n-3) + 6n(n-1)(n-2) + 7n(n-1) + n$$ · 3 years, 6 months ago

Great! Thanks. · 3 years, 6 months ago

Why did you change the limits? It was from $$\displaystyle n=1$$..you made it $$\displaystyle n=2$$.. · 3 years, 6 months ago

Because $$\frac {n-1}{(n-1)!} = 0$$ when $$n=1$$ · 3 years, 6 months ago

Oh, right..sorry. · 3 years, 6 months ago

e^x =1+ x + x^2/2.1 + x^3/3.2.1 + ..... up to infinity =1 + 2x/2.1+ 3x^2/3.2.1 + 4x^3/4.3.2.1 +.....

x.e^x = x + 2x^2/2.1 + 3x^3/3.2.1 + 4x^4/4.3.2.1 + ....... ------- (1)

differentiating eq. (1) w.r.t. x we get the above series. Then substituting x=1 we get the answer as 2e . · 3 years, 6 months ago