A factorial tail after reindexing · source reference

Printed question

It is given that $\sum_{r=1}^n\frac{r^2+3r+1}{(r+2)!}=\frac32-\frac1{(n+1)!}-\frac1{(n+2)!}$.

(a) Explain why its infinite series converges, and state the limit. [2]

(b) Find exactly $\sum_{r=6}^{\infty}\frac{r^2-r-1}{r!}$. [3]

Printed working

(a) As $n\to\infty$, $\frac1{(n+1)!}\to0,\frac1{(n+2)!}\to0$. Thus

$$\sum_{r=1}^{\infty}\frac{r^2+3r+1}{(r+2)!}=\frac32-0-0=\frac32.$$

Since it is finite, the series converges.

(b) Replace $r$ by $r+2$:

$$\begin{aligned}\sum_{r=6}^{\infty}\frac{r^2-r-1}{r!}&=\sum_{r=4}^{\infty}\frac{(r+2)^2-(r+2)-1}{(r+2)!}\\&=\sum_{r=4}^{\infty}\frac{r^2+3r+1}{(r+2)!}\\&=\sum_{r=1}^{\infty}\frac{r^2+3r+1}{(r+2)!}-\sum_{r=1}^3\frac{r^2+3r+1}{(r+2)!}\\&=\frac32-\left(\frac32-\frac1{4!}-\frac1{5!}\right)=\frac1{20}.\end{aligned}$$

Printed answer key

(a) $3/2$; (b) $1/20$

Mathematical review and qualifications

(a) Both factorial reciprocals tend to zero, so the partial sums tend to the finite value $3/2$.

(b) With $r=s+2$, $(r^2-r-1)/r!=(s^2+3s+1)/(s+2)!$. Thus the sum is $S_\infty-S_3=3/2-[3/2-1/4!-1/5!]=1/24+1/120=1/20$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [modified 2024/EJC/Prelim/I/6] Question PDF pages 2; solution PDF pages 7. The original files remain private QA records outside this native content.