Limits, Continuity & Differentiability
Riemann sums
Grade 12
Question:
<p>The value of $\lim_{n \to \infty} \left(\frac{1}{na} + \frac{1}{na+1} + \frac{1}{na+2} + \ldots + \frac{1}{nb}\right)$ is</p>
<p>(a) $\log\left(\frac{b}{a}\right)$</p>
<p>(b) $\log\left(\frac{a}{b}\right)$</p>
<p>(c) $\log(ab)$</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Recognize the sum as a Riemann sum approximating the integral of $\frac{1}{x}$ from $a$ to $b$.
<p>This is a Riemann sum. We can write the sum as $\sum_{k=0}^{n(b-a)} \frac{1}{na+k} = \frac{1}{n}\sum_{k=0}^{n(b-a)} \frac{1}{a + k/n}$.</p><p>As $n \to \infty$, this Riemann sum converges to $\int_a^b \frac{1}{x} dx = \ln b - \ln a = \log\left(\frac{b}{a}\right)$.</p>
Correct Answer: A