Finding limits using definite integrals $\lim_{n\to\infty}\sum^n_{k=1}\frac{k^{4}}{n^{5}}$

Hint: rewrite the sum as: $$\frac1n\sum_{k=1}^n\Bigl(\frac kn\Bigr)^4.$$ This is an upper Riemann sum for the function $x^4$ on the interval $[0,1]$.

Draw the curve $y=x^4$. By area comparison, we have $$\int_0^n x^4\,dx\lt 1^4+2^4+\cdots+n^4\lt \int_0^{n+1} x^4\,dx.$$ It follows that $$\frac{1}{5}\lt \frac{1^4+2^4+\cdots +n^4}{n^5}\lt \frac{1}{5}\cdot \frac{(n+1)^5}{n^5}.$$ Since $\lim_{n\to\infty}\frac{(n+1)^5}{n^5}=1$, it follows by Squeezing that our limit is $\frac{1}{5}$.

Hint:

Summation of the series using definite integral:

$$\lim \limits_{n\to \infty }\frac{1}{n} \sum \limits^{h(n)}_{r=g(n)}f(\frac{r}{n})=\int \limits^{b}_{a}f(x)dx$$

Where

1.$$\sum \to \int$$

2.$$\frac{r}{n} \to x$$

3.$$\frac{1}{n} \to dx$$

4.$$a=\lim \limits_{n\to \infty }\frac{g(n)}{n}$$

5.$$b=\lim \limits_{n\to \infty }\frac{h(n)}{n}$$