<s>[INST] Express a sum as a definite integral 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.
 [/INST] 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
$$\lim_{n\to\infty} \sum_{i=1}^n  \frac{2}{n}\left(1+ \frac{2i-1}{n}\right)^\frac{1}{3}=\lim_{n\to\infty} \frac{2}{n}\sum_{i=1}^n  \left(1+ \frac{2i}{n}\right)^\frac{1}{3}=\int_0^2 (1+x)^\frac13\ \mathsf dx$$
</s>