Limits, Continuity & Differentiability
Limit of Series Sum — Riemann Sum
nta_pyq_2024_apr
Grade 12
Question:
$\displaystyle\lim_{n\to\infty}\frac{(1^2-1)(n-1)+(2^2-2)(n-2)+\cdots+((n-1)^2-(n-1))\cdot1}{(1^3+2^3+\cdots+n^3)-(1^2+2^2+\cdots+n^2)}$ is equal to
$\dfrac{2}{3}$
$\dfrac{1}{3}$
$\dfrac{3}{4}$
$\dfrac{1}{2}$
Step-by-Step Solution
Key Concept: Numerator $=\sum_{r=1}^{n-1}(r^2-r)(n-r)=\sum_{r=1}^{n-1}(-r^3+r^2(n+1)-nr)$. Denominator $=\sum r^3-\sum r^2$. Express both in closed form and find the limit.
$\lim_{n\to\infty}\frac{(n-1)(n^2+5n-8)}{(n+1)(3n^2-n-2)}=\frac{1}{3}$.
Correct Answer: 2