First slide

Introduction to definite integration

Question

$lim_{n \to \infty} (\frac{1}{n} + \frac{1}{n + 1} + \dots + \frac{1}{3 n}) =$

Easy

A

log 2
B

log 3
C

log 5
D

0

Solution

$\begin{array}{l} lim_{n \to \infty} (\frac{1}{n} + \frac{1}{n + 1} + \dots + \frac{1}{3 n}) \\ = lim_{n \to \infty} (\frac{1}{n} + \frac{1}{n + 1} + \dots + \frac{1}{n + 2 n}) \\ = lim_{n \to \infty} \sum_{k = 0}^{2 n} \frac{1}{n + k} = lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{2 n} \frac{1}{1 + k / n} \end{array}$
$= \int_{0}^{2} \frac{d x}{1 + x} = {\log (1 + x)|}_{0}^{2} = \log 3$

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