How to calculate $\lim_{n \to \infty} \frac 1{3n} +\frac 1{3n+1}+\cdots+\frac 1{4n}$?

Question

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 1

Hint: Represent this expression as a Riemann sum: $$\frac{1}{n}\sum_{k=0}^{n}\frac{1}{3+\frac{k}{n}}\begin{array}{c}{_{n\rightarrow\infty}\\ \longrightarrow\\}\end{array} \int_0^1\frac{dx}{3+x}=\ln\frac43.$$

Answer 2

Alternative solution: use that $$\exists \lim_{n\to\infty}\left({1+\frac12+\cdots+\frac1n-\ln n}\right).$$