Summation and series: ACJC 2024 Promo Q8

(a)

(i) $$\begin{aligned} \sum_{r=1}^{n}\frac{1}{4r^2 - 1} &= \frac{n}{2n + 1}\\ &= \frac{1}{2} - \frac{1}{2(2n + 1)} \end{aligned}$$ As $n\to\infty$, $\dfrac{1}{2(2n+1)}\to0$. $$\begin{aligned} \sum_{r=1}^{\infty}\frac{1}{4r^2 - 1} &= \frac{1}{2} \end{aligned}$$

Alternative: $$\begin{aligned} \sum_{r=1}^{\infty}\frac{1}{4r^2-1} &=\lim_{n\to\infty}\frac{n}{2n+1}\\ &=\lim_{n\to\infty}\frac{1}{2+\frac1n}\\ &=\frac12 \end{aligned}$$ Since $\dfrac12$ is a constant, the series converges.

(a)

(ii) Replacing $r$ with $r - 1$: $$\begin{aligned} \sum_{r=6}^{N}\frac{1}{(2r + 1)(2r + 3)} &= \sum_{r-1=6}^{r-1=N}\frac{1}{(2(r-1)+1)(2(r-1)+3)}\\ &= \sum_{r=7}^{N+1}\frac{1}{(2r - 1)(2r + 1)}\\ &= \sum_{r=1}^{N+1}\frac{1}{4r^2 - 1} - \sum_{r=1}^{6}\frac{1}{4r^2 - 1}\\ &= \frac{N + 1}{2(N + 1) + 1} - \frac{6}{12 + 1}\\ &= \frac{N + 1}{2N + 3} - \frac{6}{13}\\ &= \frac{N - 5}{13(2N + 3)}. \end{aligned}$$

(b)

$$\begin{aligned} u_2 &= 1 - \frac{1}{2} = \frac{1}{2}\\ u_3 &= 1 - \frac{1}{\frac{1}{2}} = -1\\ u_4 &= 1 - \frac{1}{-1} = 2 \end{aligned}$$

Alternative: Use GC $$\begin{aligned} \sum_{r=1}^{50}u_r&=2\times17+\frac12\times17+(-1)\times16\\ &=26.5 \end{aligned}$$