Sum it again

Calculus Level 3

$\large 1+\frac{2^{2}}{2!}+\frac{3^{2}}{3!}+\frac{4^{2}}{4!}+\ldots$

If the series above equals to $n\times e$ for $\displaystyle e =\lim_{x\to0} \left(1+\frac1x\right)^x$ , find the value of $n$ .

The answer is 2.

2 solutions

Curtis Clement
Aug 7, 2015

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

thank u for getting me rid out of this problem

sumit cooooool - 5 years, 10 months ago

Michael Mendrin
Aug 7, 2015

For this one, make use of the fact that

$\dfrac { { n }^{ 2 }-n }{ n! } =\dfrac { 1 }{ (n-2)! }$

Sum it again

The answer is 2.

2 solutions

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