Limits, Continuity & Differentiability
Limit as Riemann Sum — Inverse Trig Result
nta_pyq_2024_apr
Grade 12
Question:
Let $\displaystyle\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2n}{(n^2+1)\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8n}{(n^2+4)\sqrt{n^4+16}}+\cdots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2n\cdot n^2}{(n^2+n^2)\sqrt{n^4+n^4}}\right)$ be $\dfrac{\pi}{k}$, using only the principal values of the inverse trigonometric functions. Then $k^2$ is equal to
Step-by-Step Solution
Key Concept: The general term is $\frac{n}{\sqrt{n^4+r^4}}-\frac{2nr^2}{(n^2+r^2)\sqrt{n^4+r^4}}=\frac{n(n^2-r^2)}{(n^2+r^2)\sqrt{n^4+r^4}}$. This converts to a Riemann integral $\int_0^1\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}dx$.
Integral $=\frac{\pi}{4\sqrt{2}}$, so $K=4\sqrt{2}$, $K^2=32$.
Correct Answer: 32