Evaluate $\lim_{N \to \infty}$ $\sum_{k=1}^N$ $k^3 \over N^4$ (Riemann sum)

Question

Given,

$\sum_{k=1}^N $$k^3 \over N^4$,

I have to evaluate

$\lim_{N \to \infty}$ $\sum_{k=1}^N$ $k^3 \over N^4$

with the help of the Riemann sum.

Approach

Assume

$a_N$ $:=$ $1 \over N$,

$x_k$ $:=$ $k \over N$,

$f(x_k)$ $:=$ $k^3 \over N^4$.

Then, I would have to start with:

$\sum_{k=1}^N$ $k^3 \over N^4$ $*$ $k \over N$ = $\sum_{k=1}^N$ $k^4 \over N^5$ = $1 \over N^5$ * $\sum_{k=1}^N k^4$,

but when I evaluate the limit, the whole sum would directly converge to zero, which can't be the solution here. So I guess I did something wrong here.

Answer 1

$$\lim_{N\to\infty}\sum_{k=1}^N\dfrac{k^r}{N^{r+1}}=\lim_{N\to\infty}\dfrac1N\sum_{k=1}^N\left(\dfrac kN\right)^r$$

Now use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

Answer 2

It is important to determine the bounds of the integral correctly. As @lab explained, your summand needs to be rewritten as $\frac{1}{n} (\frac{k}{n})^3$, but it is in fact $\frac{1}{n} ( 0 + (1-0) \times (\frac{k}{n})^3$), with $a=0, b=1$ being lower and upper bounds of the integral.