What is the Sum of Factorials for a Specific Equation?

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SUMMARY

The discussion evaluates the sum of factorials for the equation $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$. It establishes that $$\dfrac{n^2-2}{n!}$$ can be rewritten as $$\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!}$$. The final result of the summation from 2 to 2012 is calculated as $$3-\dfrac{1}{2011!}-\dfrac{2}{2012!}$$, confirming the convergence of the series to a specific numerical value.

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Evaluate $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$
 
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anemone said:
Evaluate $$\frac{2^2-2}{2!}+\frac{3^2-2}{3!}+\frac{4^2-2}{4!}+\cdots+\frac{2012^2-2}{2012!}$$
$\dfrac{n^2-2}{n!}=\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!}$

$\therefore \sum_{2}^{2012}(\dfrac{n^2-2}{n!})=$$ \sum_{2}^{2012}(\dfrac{1}{(n-1)!}-\dfrac{1}{n!}+\dfrac{1}{(n-2)!}-\dfrac{1}{n!})$

$=3-\dfrac {1}{2011!}-\dfrac{2}{2012!}$
 

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