Series Demystified 7

$n^3-3n-1$

$=n^3-n-2n-2+1$

$=n(n^2-1)-2(n+1)+1$

$=(n-1)(n)(n+1)-2(n+1)+1$

$\displaystyle\sum_{n=1}^{48}\dfrac{n^3-3n-1}{(n+1)!}$

$=\displaystyle\sum_{n=1}^{48}\frac{(n-1)(n)(n+1)-2(n+1)+1}{(n+1)!}$

$=\displaystyle\sum_{n=1}^{48}(\frac{1}{(n-2)!}-\frac{2}{n!}+\frac{1}{(n+1)!})$

$=\displaystyle\sum_{n=1}^1\frac{n^3-3n-1}{(n+1)!}+\sum_{n=2}^{48}(\frac{1}{(n-2)!}-\frac{2}{n!}+\frac{1}{(n+1)!})$

$=-\displaystyle\frac{3}{2}+\frac{1}{0!}+\frac{1}{1!}-\frac{1}{2!}-\frac{1}{47!}-\frac{1}{48!}+\frac{1}{49!}$

$=\dfrac{-48×49-49+1}{49!}$

$=\dfrac{-2400}{49!}$

Thus the answer is $\boxed{-2400}$ .

$\begin{aligned} S &= \sum_{n=1}^{48} \frac{n^3-3n-1}{(n+1)!} \\ &= \sum_{n=1}^{48} \frac{(n^3+n^2-n-1)-(n^2+2n)}{(n+1)!} \\ &= \sum_{n=1}^{48} \frac{(n-1)(n+1)^2}{(n+1)!} - \frac{n(n+2)}{(n+1)!} \\ &= \sum_{n=1}^{48} \frac{(n-1)(n+1)}{n!} - \frac{n(n+2)}{(n+1)!} \\ &= \frac{(1-1)(1+1)}{1!} - \frac{(48)(48+2)}{(48+1)!} & \text{since this is a telescoping series} \\ &= -\frac{2400}{49!} \end{aligned}$

Hence $49! \times S = \boxed{-2400}$ .

$n^3-3n-1=(n-1)n(n+1)-2n(n+1)+1.\\ \therefore~\displaystyle S= \sum_{n=1}^{48}\dfrac{n^3-3n-1}{(n+1)!} = \sum_{n=1}^{48} \dfrac{(n-1)n(n+1)-2n(n+1)+1}{(n+1)!} =\sum_{n=1}^{48}\Big\{\dfrac 1 {(n-2)!} - \dfrac 2 {n!}+\dfrac 1 {(n+1)!} \Big \}.\\ S~is~ a~ mix~ telescopic~ series.~First~term~for~n =1~is~\dfrac 1 {(-1)!},~but~~negative~factorial~is~undefined. \\ \therefore~Original~expression~is~used~to~calculate~n=1. ~~~ n=2~to~n=48~is~the~ telescopic~ series.\\ \therefore~\displaystyle S= \sum_{n=1}^{48}\dfrac{n^3-3n-1}{(n+1)!} = \sum_{n=1}^1 \dfrac{n^3-3n-1}{(n+1)!} +\sum_{n=2}^{48}\Big \{\dfrac 1 {(n-2)!} - \dfrac 2 {n!}+\dfrac 1 {(n.+1)!} \Big \}\\ \displaystyle =-\dfrac 3 2+\underline { \sum_{n=0}^{46} \dfrac 1{n!}-\sum_{n=2}^{48} \dfrac 1 {n!}} -\underline{\sum_{n=2}^{48} \dfrac 1 {n!}+\sum_{n=3}^{49} \dfrac 1 {n!}} =-\dfrac 3 2+\underline { \dfrac 1{0!}+\dfrac 1{1!}-\dfrac 1{47!}- \dfrac 1 {48!}} -\underline{ \dfrac 1 {2!}+ \dfrac 1 {49!}} = -\dfrac 1{47!}- \dfrac 1 {48!}+\dfrac 1 {49!}\\ So~(49!*S)=-\dfrac {49!}{47!}- \dfrac {49!} {48!}+\dfrac {49!} {49!}= - 49*48 - 49 + 1=-49^2+1=\Large~~~\color{#D61F06}{-2400}.$