Express a sum as a definite integral

Question

I need help expressing $\lim_{n\to\infty} \sum_{i=1}^n \frac{2}{n}(1+ \frac{2i-1}{n})^\frac{1}{3} $ as a definite integral. I am usually pretty good with figuring out these problems but I don't know what to do with the $-1$ term in the numerator and I can't find similar examples online.

Answer 1

We have that

$$\left(1+ \frac{2i}{n}\right)^\frac{1}{3}-\left(1+ \frac{2i-1}{n}\right)^\frac{1}{3}=\frac{\frac1n}{\left(1+ \frac{2i}{n}\right)^\frac{2}{3}+\left(1+ \frac{2i}{n}\right)^\frac{1}{3}\left(1+ \frac{2i-1}{n}\right)^\frac{1}{3}+\left(1+ \frac{2i}{n}\right)^\frac{2}{3}}\le$$

$$\le \frac1n\frac{1}{3\left(1+ \frac{2i}{n}\right)^\frac{2}{3}}$$

and

$$\lim_{n\to\infty} \left(\frac{2}{n} \cdot \frac1n \sum_{i=1}^n \frac{1}{3\left(1+ \frac{2i}{n}\right)^\frac{2}{3}}\right)=0$$

therefore

$$\lim_{n\to\infty} \sum_{i=1}^n \frac{2}{n}\left(1+ \frac{2i-1}{n}\right)^\frac{1}{3}=\lim_{n\to\infty} \frac{2}{n}\sum_{i=1}^n \left(1+ \frac{2i}{n}\right)^\frac{1}{3}=\int_0^2 (1+x)^\frac13\ \mathsf dx$$

Answer 2

$$\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\frac{2}{n} (1+\frac{2k-1}{n})^{1/3}= \int_{0}^{1} 2 (1+2x)^{1/3} dx= \frac{3}{4} (3^{4/3}-1).$$ Here we have used $k/n=x$.