How to do this limit: $\lim\limits_{n\to \infty}\large\frac{\sum_{k=1}^n k^p}{n^{p+1}}$?

Question

$$\large\lim_{n\to \infty}\large\frac{\sum_{k=1}^n k^p}{n^{p+1}}$$

I'm stuck here because the sum is like this: $1^p+2^p+3^p+4^p+\cdots+n^p$.

Any ideas?

Answer 1

Hint: Recall that if $f$ is integrable on $[a,b]$, then:

$$ \int_a^b f(x)~dx = \lim_{n\to \infty} \dfrac{b-a}{n}\sum_{k=1}^n f \left(a + k \left(\dfrac{b-a}{n}\right) \right) $$

Can you rewrite the given sum in the above form? What might be an appropriate choice for $a$, $b$, and $f(x)$?

Answer 2

By Writing limit-of-sum as Integral we get

$$ \displaystyle\lim_{n\to \infty}\large\frac{\sum_{k=1}^n k^p}{n^{p+1}}=\lim_{n\to \infty} \sum_{k=1}^n \frac{k^p}{n\cdot n^p}=\lim_{n\to \infty} \dfrac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^p=\int_0^1 x^p dx=\frac{1}{p+1}$$

Answer 3

HINT: This hint is probably a hammer but still worth it. Faulhaber's formula states that $$\sum_{k=1}^n k^p = \dfrac1{p+1} \sum_{j=0}^p (-1)^j \dbinom{p+1}j B_j n^{p+1-j}$$