AIEEE 2005 (Offline)Question 1
$\lim\limits_{n\to\infty}\left[\frac{1}{n^{2}}\sec^{2}\frac{1}{n^{2}}+\frac{2}{n^{2}}\sec^{2}\frac{4}{n^{2}}+\cdots+\frac1n\sec^{2}1\right]$ equals
Definite Integration
Solution
Concept. The $r$-th term is $\frac{r}{n^2}\sec^2\frac{r^2}{n^2}=\frac1n\cdot\frac rn\sec^2\big(\frac rn\big)^2$, a Riemann sum of $x\sec^2(x^2)$ on $[0,1]$. Formulas used. (1) $\lim\frac1n\sum_{r=1}^n\phi\big(\frac rn\big)=\int_0^1\phi(x)\,dx$(2) $\int x\sec^2(x^2)\,dx=\frac12\tan(x^2)$ (put $t=x^2$)Step 1 - General term. The last term $\frac1n\sec^21$ is the case $r=n$, so the sum is $\frac1n\sum_{r=1}^n\frac rn\sec^2\big(\frac rn\big)^2$. Step 2 - Use (1), (2). $$\int_0^1x\sec^2(x^2)\,dx=\Big[\frac12\tan(x^2)\Big]_0^1=\frac12\tan1.$$ Answer: (D)