Could you please help me calculate this limit:
$\lim_{n \to \infty} \frac 1{3n} +\frac 1{3n+1}+\cdots+\frac 1{4n}$.
My best try is :
$\lim_{n \to \infty} \frac 1{3n} +\frac 1{3n+1}+\cdots+\frac 1{4n}=\lim_{n \to \infty}\sum_{k=3n}^{4n}\frac 1n$
$\frac 14 \leftarrow \frac{n+1}{4n}\le \sum_{k=3n}^{4n}\frac 1n \le \frac{n+1}{3n} \to \frac 13$.
Thanks.
Answer
Hint: Represent this expression as a Riemann sum:
$$\frac{1}{n}\sum_{k=0}^{n}\frac{1}{3+\frac{k}{n}}\begin{array}{c}{_{n\rightarrow\infty}\\ \longrightarrow\\}\end{array} \int_0^1\frac{dx}{3+x}=\ln\frac43.$$
No comments:
Post a Comment