Sequences & Series
Limit of Partial Sum Using Telescoping
nta_pyq_2025_apr
Grade 11
Question:
If $\displaystyle\sum_{r=1}^n T_r = \dfrac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\displaystyle\lim_{n\to\infty}\sum_{r=1}^n\dfrac{1}{T_r}$ equals
0
$\dfrac{2}{3}$
1
$\dfrac{1}{3}$
Step-by-Step Solution
Key Concept: Find $T_n=S_n-S_{n-1}$ and simplify to a product of three consecutive odd numbers, then use partial fractions to telescope $\sum 1/T_r$.
$T_n=S_n-S_{n-1}$. Substituting and simplifying (the product $(2n-1)(2n+1)(2n+3)(2n+5)-(2n-3)(2n-1)(2n+1)(2n+3)$ factors as $8(2n-1)(2n+1)(2n+3)$):
$$T_n=\frac{1}{8}(2n-1)(2n+1)(2n+3).$$
Partial fraction: $\dfrac{1}{T_n}=\dfrac{8}{(2n-1)(2n+1)(2n+3)}=\dfrac{2}{(2n-1)(2n+1)}-\dfrac{2}{(2n+1)(2n+3)}$.
This telescopes:
$$\sum_{r=1}^n\frac{1}{T_r}=2\left(\frac{1}{1\cdot3}-\frac{1}{(2n+1)(2n+3)}\right)\xrightarrow{n\to\infty}\frac{2}{3}.$$
Correct Answer: 2