An error bound and a shifted rational sum · source reference

Printed question

The sum $\sum_{r=1}^n\frac4{(2r+3)(2r+5)}$ is denoted $S_n$. Given $S_n=\frac25-\frac2{2n+5}$,

(i) find the smallest $n$ for which $S_n$ is within $10^{-3}$ of the sum to infinity [3];

(ii) find $\sum_{r=10}^{2n}\frac2{4r^2-1}$ in terms of $n$. [3]

Printed working

(i)

$$|S_n-S_{\infty}|<10^{-3}\Rightarrow\left|\left(\frac25-\frac2{2n+5}\right)-\frac25\right|<10^{-3}\Rightarrow\frac2{2n+5}<10^{-3}.$$

$2n+5>2000\Rightarrow n>997.5$. Smallest $n=998$.

(ii)

$$\begin{aligned}\sum_{r=10}^{2n}\frac2{4r^2-1}&=\frac12\sum_{r=10}^{2n}\frac4{(2r-1)(2r+1)}\\&=\frac12\sum_{r=8}^{2n-2}\frac4{[2(r+2)-1][2(r+2)+1]}\\&=\frac12\sum_{r=8}^{2n-2}\frac4{(2r+3)(2r+5)}\\&=\frac12\left[\sum_{r=1}^{2n-2}\frac4{(2r+3)(2r+5)}-\sum_{r=1}^7\frac4{(2r+3)(2r+5)}\right]\\&=\frac12\left[\left(\frac25-\frac2{2(2n-2)+5}\right)-\left(\frac25-\frac2{2(7)+5}\right)\right]\\&=\frac1{19}-\frac1{4n+1}.\end{aligned}$$

Teacher notes

“Within” is strictly “$<$” and does not include “$=$”.

Factorise $4r^2-1$ and compare $(2r-1)(2r+1)$ with $(2r+3)(2r+5)$ to find the replacement for $r$. Replace $r$ by $r+2$.

To use the given partial-sum result, the lower limit must be 1: $\sum_{r=8}^{2n-2}f(r)=\sum_{r=1}^{2n-2}f(r)-\sum_{r=1}^7f(r)$.

Printed answer key

(i) 998; (ii) $1/19-1/(4n+1)$

Mathematical review and qualifications

(i) $S_\infty=2/5$. The condition $2/(2n+5)<10^{-3}$ gives $n>997.5$, hence $n=998$.

(ii) The reciprocal differences cancel, leaving $1/19-1/(4n+1)$ for integer $n\ge5$. Equivalently this is $\frac12(S_{2n-2}-S_7)$ after replacing $r$ by $r-2$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [modified 2020/ASRJC/Prelim/I/5] Question PDF pages 2; solution PDF pages 8. The original files remain private QA records outside this native content.