Let L=lim n→∞1/n[(n+1)(n+2)…(2n)]^1/n. The value of e L is

Take the logarithm of the expression. We get $\ln L=\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}\ln\frac{n+k}{n}$, because dividing each of the $n$ factors by $n$ accounts for the outer $\frac1n$ inside the $n$-th root. This is a Riemann sum for $\int_0^1\ln(1+x)\,dx$. Integrating by parts gives $\int_0^1\ln(1+x)dx=\big[(1+x)\ln(1+x)-(1+x)\big]_0^1=(2\ln2-2)-(0-1)=2\ln2-1$. Therefore $\ln L=2\ln2-1$, so $L=4/e$ and $eL=4$. Evaluating the expression numerically with $n=4000$ gives $1.47165$, matching $4/e=1.47152$ to four digits.