The Riemann sum-to-integral conversion is a standard JEE technique for evaluating limits of the form lim sum.

Standard Form: lim(n->inf) $\frac{1}{n}$ * sum(r=1 to n) f$\frac{r}{n}$ = integral(0 to 1) f(x) dx

Generalized Form: lim(n->inf) $\frac{1}{n}$ * sum(r=a to bn) f$\frac{r}{n}$ = integral(a/n... -> 0 to b) ...

More precisely: if the sum goes from r = an to r = bn, replace r/n by x, and limits become a to b.

Common Examples:

1. lim [1/n * $\frac{1+2+...+n}{n}$] = integral(0,1) x dx = 1/2
2. lim [1^k+2^k+...+$n^k$]/n^(k+1) = integral(0,1) $x^k$ dx = $\frac{1}{k+1}$
3. lim [$\frac{1}{n+1}$+...+1/2n] = integral(0,1) $\frac{1}{1+x}$ dx = ln 2
4. lim [(n!)^$\frac{1/n}{n}$] = e^(-1) (using integral(0,1) lnx dx = -1)
5. lim [(2n choose n)]^$\frac{1}{n}$ = 4 (using integral(0,1) [ln x + ln(1-x)] dx)

Strategy:

1. Factor out 1/n from the sum
2. Express each term as $\frac{1}{n}$*f$\frac{r}{n}$
3. Identify f and the limits
4. Write the integral
5. Evaluate

Pitfall: When the sum has unusual limits (e.g., r=1 to 3n), the integral limits adjust: r/n goes from 1/n->0 to 3n/n=3, giving integral(0 to 3).