Let S _ n = 1 · ( n - 1 ) + 2 · ( n - 2 ) + 3 · ( n - 3 ) + + ( n -...

\text{Let}{S}_{n} = 1 \cdot ( n - 1 ) + 2 \cdot ( n - 2 ) + 3 \cdot ( n - 3 ) + \ldots + ( n - 1 ) \cdot 1, n ⩾ 4 . Generalformof{S}_{n} = \mathop{∑}_{r = 1}^{n - 1}r ( n - r ) = \dfrac{n\left({n}^{2} - 1\right)}{6}$</p><p>$\dfrac{2{S}_{n}}{n !} = \dfrac{2n\left(n + 1\right)\left(n - 1\right)}{6n\left(n - 1\right)\left(n - 2\right) !} = \dfrac{( n + 1 )}{3 ( n - 2 ) !}$</p><p>$⇒ \mathop{∑}_{n = 4}^{\infty}\left(\dfrac{2{S}_{n}}{n !} - \dfrac{1}{( n - 2 ) !}\right)$⇒ \mathop{∑}_{n = 4}^{\infty}\left(\dfrac{( n + 1 )}{3 ( n - 2 ) !} - \dfrac{1}{( n - 2 ) !}\right)$

$= \mathop{∑}_{n = 4}^{\infty}\dfrac{( n - 2 )}{3 ( n - 2 ) !}$

$= \dfrac{1}{3}\mathop{∑}_{n = 4}^{\infty}\dfrac{1}{( n - 3 ) !}$

$= \dfrac{1}{3}\left(\dfrac{1}{1 !} + \dfrac{1}{2 !} + \dfrac{1}{3 !} + \ldots . .\right) = \dfrac{e - 1}{3}$