$\lim_{n\to\infty}\sum_{i=1}^{4n-1}(\frac{i}{n})^3\frac{1}{n}$

Question

What is the result of the following limit?

$$\lim_{n\to\infty}\sum_{i=1}^{4n-1}(\frac{i}{n})^3\frac{1}{n}$$

I see two approaches here, either doing it as a Riemann sum, then I get $4^4\int_0^1 x^3dx$, or taking $\lim_{n\to\infty}(\frac{(4n-1)(4n)}{2})^2\frac{1}{n^4}$.

Both results in $64$. But according to the book where I found the problem, the correct result is $16$.

Am I doing something wrong, or is there an error in the book?

Answer 1

Since $$\sum_{i=1}^{4n-1}\left(\frac in\right)^3\frac 1n=4^4\frac 1{4n}\sum_{i=1}^{4n-1}\left(\frac i{4n}\right)^3\to 4^4\int_0^1x^3\mathrm dx=4^3=64$$ (the convergence holds because $a_n:=\frac 1n\sum_{i=1}^n\left(\frac in\right)^3\to\int_0^1x^3\mathrm dx$ hence the sub-sequence $(a_{4n})_{n\geqslant 1}$ converges to the same limit).

Answer 2

You can also write $$S(n)=\sum_{i=1}^{4n-1}\left(\frac in\right)^3\frac 1n=\frac{1}{n^4}\sum_{i=1}^{4n-1}i^3$$ and use Faulhaber formula to get $$\sum_{i=1}^{4n-1}i^3=4 n^2 (4 n-1)^2$$ and then $$S(n)=\frac{4 (4 n-1)^2}{n^2}$$