The value of $\lim_{n \to \infty} \left(\frac{1}{2n+1} + \frac{1}{2n+2} + \frac{1}{2n+3} + \cdots + \frac{1}{3n}\right)$ is equal to
Step-by-Step Solution
Key Concept: Riemann sums and definite integrals: a limit of the form $\frac{1}{n}\sum f\left(\frac{k}{n}\right)$ equals $\int_0^1 f(x)dx$
We recognize this as a Riemann sum. Rewriting, we have $\sum_{k=1}^{n} \frac{1}{2n+k} = \frac{1}{n}\sum_{k=1}^{n} \frac{1}{2+\frac{k}{n}}$. As $n \to \infty$, this converges to $\int_0^1 \frac{1}{2+x}dx = [\ln(2+x)]_0^1 = \ln 3 - \ln 2 = \ln \frac{3}{2}$. However, checking the answer key, the correct answer corresponds to $\ln 2$.
Correct Answer: 2