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 \(n^4 = n(n-1)(n-2)(n-3) + 6n(n-1)(n-2) + 7n(n-1) + n \)

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

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 \(n^4 = n(n-1)(n-2)(n-3) + 6n(n-1)(n-2) + 7n(n-1) + n \)

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Great! Thanks.

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Why did you change the limits? It was from \(\displaystyle n=1\)..you made it \(\displaystyle n=2\)..

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Because \( \frac {n-1}{(n-1)!} = 0 \) when \(n=1 \)

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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 .

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Nicely done!

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