Harmonic sum In Chapter 10, we will encounter the harmonic sum 1 + 1/2 + 1/3 + ⋯ + 1/n. Use a left Riemann sum to approximate ∫[1 to n+1] (dx/x) (with unit spacing between the grid points) to show that 1 + 1/2 + 1/3 + ⋯ + 1/n > ln(n + 1). Use this fact to conclude that lim (n → ∞) (1 + 1/2 + 1/3 + ⋯ + 1/n) does not exist.

Verified step by step guidance

1. Recall that the integral \( \int_1^{n+1} \frac{1}{x} \, dx \) represents the area under the curve \( y = \frac{1}{x} \) from \( x=1 \) to \( x=n+1 \). This integral evaluates to \( \ln(n+1) \).
2. Set up a left Riemann sum with unit spacing to approximate the integral \( \int_1^{n+1} \frac{1}{x} \, dx \). The left endpoints for the subintervals are \( x = 1, 2, 3, \ldots, n \), so the sum is \( \sum_{k=1}^n \frac{1}{k} \cdot 1 = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \).
3. Since \( \frac{1}{x} \) is a decreasing function, the left Riemann sum overestimates the integral. Therefore, we have the inequality \( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} > \int_1^{n+1} \frac{1}{x} \, dx = \ln(n+1) \).
4. Use this inequality to analyze the behavior of the harmonic sum as \( n \to \infty \). Because \( \ln(n+1) \to \infty \) as \( n \to \infty \), the harmonic sum \( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \) must also grow without bound.
5. Conclude that the limit \( \lim_{n \to \infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right) \) does not exist as a finite number, since the harmonic sum diverges to infinity.

Key Concepts

Harmonic Series

The harmonic series is the sum of reciprocals of natural numbers: 1 + 1/2 + 1/3 + ... + 1/n. It grows without bound as n increases, but very slowly. Understanding its divergence is key to analyzing its behavior compared to integrals and logarithmic functions.

Left Riemann Sum Approximation

A left Riemann sum approximates the integral of a function by summing the areas of rectangles using the function's value at the left endpoint of each subinterval. For f(x) = 1/x on [1, n+1] with unit spacing, this sum corresponds to the harmonic sum, linking discrete sums to continuous integrals.

Improper Integral and Limit Comparison

The integral ∫[1 to n+1] (1/x) dx equals ln(n+1), which grows without bound as n → ∞. Comparing the harmonic sum to this integral shows the sum exceeds ln(n+1), proving the harmonic series diverges and its limit does not exist (it tends to infinity).