An infinite tail from a supplied sum · source reference

Printed question

Given $\sum_{r=1}^n\frac{r+2}{2^{r+1}r(r+1)}=\frac12-\frac1{(n+1)2^{n+1}}$, evaluate $\sum_{r=5}^{\infty}\frac{r+2}{2^{r+1}r(r+1)}$, leaving the answer as $p/q$ where $p,q$ are real. [4]

Printed working

$$\begin{aligned}\sum_{r=5}^{\infty}\frac{r+2}{2^{r+1}r(r+1)}&=\sum_{r=1}^{\infty}\frac{r+2}{2^{r+1}r(r+1)}-\sum_{r=1}^4\frac{r+2}{2^{r+1}r(r+1)}\\&=\lim_{n\to\infty}\left[\frac12-\frac1{(n+1)2^{n+1}}\right]-\left[\frac12-\frac1{5(2^5)}\right]\\&=(\frac12-0)-(\frac12-\frac1{5(2^5)})=\frac1{160}.\end{aligned}$$

Teacher notes

To use the given result (lower limit $r=1$), re-express $\sum_{r=5}^{\infty}f(r)=\sum_{r=1}^{\infty}f(r)-\sum_{r=1}^4f(r)$. Note $\sum_{r=1}^{\infty}f(r)=\lim_{n\to\infty}\sum_{r=1}^nf(r)$.

Printed answer key

$1/160$

Mathematical review and qualifications

$S_\infty=1/2$ because $1/[(n+1)2^{n+1}]\to0$. Hence the requested tail is $S_\infty-S_4=1/2-[1/2-1/(5\cdot2^5)]=1/160$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [modified 2008/TPJC/Prelim/I/7] Question PDF pages 2; solution PDF pages 6, 7. The original files remain private QA records outside this native content.