Show that for all positive intergers $k$, $1^k+2^k+\dots+n^k=\Theta(n^{k+1})$

Question

Show that for all positive intergers $k$, $1^k+2^k+\dots+n^k=\Theta(n^{k+1})$. I know that this is equivalent to showing that $\liminf_{n\to\infty}\frac{\sum_{j=1}^nj^k}{n^{k+1}}>0$ and $\limsup_{n\to\infty}\frac{\sum_{j=1}^nj^k}{n^{k+1}}<\infty$. The second statement is easy to prove, but doesn't $$\liminf_{n\to\infty}\frac{1^k+2^k+\dots+n^k}{n^{n+1}}=0,$$ rendering the statement false?

Answer 1

For all $n$ and $k\in\mathbb N$, we have the following:

$$\frac{\sum_{j=1}^nj^k}{n^{k+1}}>\frac{\int_1^nx^k\ dx}{n^{k+1}}=\frac{\frac1{k+1}\left(n^{k+1}-1\right)}{n^{k+1}}=\frac1{k+1}\left(1-\frac1{n^{k+1}}\right)\ge0$$

One can similarly create an upper bound and apply the squeeze theorem to show that

$$\lim_{n\to\infty}\frac{\sum_{j=1}^nj^k}{n^{k+1}}=\frac1{k+1}$$

which provides the exact value for $\liminf$ and $\limsup$.

Answer 2

One can use Cezaro-Stolz theorem, to deduce the limit immediately.

There is another way to derive it using Bernoulli numbers. This method essentially proves that the sum $$1^k+2^k+3^k+...n^k$$ is a polynomial of degree $k+1$ with leading coefficient $\dfrac{1}{k+1}$.