Answer to Question #275105 in Real Analysis for Mao

Question

Evaluate

lim 3nΣr=1 n^2/(4n+r)^3

n→∞

Answer 1

$\displaystyle\text{Given } \lim_{n \rightarrow \infty} \sum_{r=1}^{3n}\frac{n^2}{(4r+n)^3} \\ \lim_{n \rightarrow \infty} \sum_{r=1}^{3n}\frac{n^2}{(4n+r)^3} = \lim_{n \rightarrow \infty} \sum_{r=1}^{3n}\frac{\frac{1}{n}}{(4+\frac{r}{n})^3} \\ \text{Using the Riemann sum put } x_{r} = \frac{r}{n} \implies x_{r+1} - x_r = \frac{1}{n} = \Delta x \\ \text{when } r=1 ; \lim_{n \rightarrow \infty} \frac{r}{n} = \lim_{n \rightarrow \infty} \frac{1}{n} = 0 \\ \text{when } r=3n ; \lim_{n \rightarrow \infty} \frac{r}{n} = \lim_{n \rightarrow \infty} \frac{3n}{n} = 3 \\ \implies \lim_{n \rightarrow \infty} \sum_{r=1}^{3n}\frac{n^2}{(4n+r)^3} = \int_0^3 \frac{dx}{(4+x)^3} = \left[\frac{-1}{2(4+x)^2}\right]_{0}^3 = \frac{-1}{98} + \frac{1}{32} = \frac{33}{1568} \\ \therefore \lim_{n \rightarrow \infty} \sum_{r=1}^{3n}\frac{n^2}{(4n+r)^3} = \frac{33}{1568}$