sorry if the question title is incorrect. I'm doing a homework regarding Big O notation.
$$ \text{Show that} \,\,\, \sum_{k=1}^{n}k^{\,j} \, \approx n^{\,j+1} $$ where $\approx$ is defined as: $ f(x) \approx g(x) \iff f(x) \in \mathcal{O}(g(x)) \text{ and } g(x) \in \mathcal{O}(f(x))$
My current approach is that I'm trying to show that (all limits are for $n$ going to infinity, it didn't display correctly when I typed):
$$ \lim \frac{\sum_{k=1}^{n}k^{\,j}}{n^{j+1}} \in \mathbb{R} $$ using two inequalities $$ \lim \frac{n^j}{n^{j+1}} \leq \lim \frac{\sum_{k=1}^{n}k^{\,j}}{n^{j+1}} \leq \text{don't know what to put here} $$ where the third limit should be a real number. Then the middle limit would also be real number and therefore $ \sum_{k=1}^{n} k^{\,j} \in \mathcal{O}(n^{j+1})$.
Analogically, I would need to also show that $$ \lim \frac{n^{j+1}}{\sum_{k=1}^{n}k^{\,j}} \in \mathbb{R} $$ and therefore $ n^{j+1} \in \mathcal{O}(\sum_{k=1}^{n}k^{j})$
But I'm having difficulty with even the first inequalities.
So my question is: is my approach correct/effective and if it is, what upper bound should I use? So far every upper bound I came up with had infinite limit which isn't useful.
HINT:
$$\frac{\sum_{k=1}^nk^j}{n^{j+1}}=\frac1n\sum_{k=1}^n\left(\frac{k}n\right)^j\;;$$
now what is the largest term in the last sum?