I stumbled upon an interesting limit of series and I was wondering how could it be solved...
$$\lim_{n\to \infty} \frac{1^p + 2^p + \ldots + n^p}{n^{p+1}} = \frac{1}{p+1}$$
I tried to divide it into smaller sections ($\frac{1^p}{n^{p+1}}$, $\frac{2^p}{n^{p+1}}$, ..., $\frac{1}{n^p}$), factoring the numerator, but nothing seemed to be the way to go...
I also tried using the squeeze theorem - finding two series with the limit of ${\frac{1}{p+1}}$ but I had no luck with this either.
Any ideas?
As Daniel Fischer suggested in the comments, rewrite your quantity as $$ F_p(n) = \frac{\left(\frac{1}{n}\right)^p+\left(\frac{2}{n}\right)^p+\dots+\left(\frac{n}{n}\right)^p}{n} = \frac{1}{n}\sum_{k=1}^n f\!\left(\frac{k}{n}\right) $$ with $f\colon x\in[0,1]\mapsto x^p$. Then, use the theorem about Riemann sums to relate the limit of $F_p(n)$ to $\int_0^1 f$.
Hint Use Stolz-Ceazaro Theorem.
If you are not familiar with Stolz-Cezaro, look at the area under the curve $f(x)=x^p$ from 1 to $n$.
By dividing $[1,n]$ into $n$ intervals of length 1, and using the rectangles with height given by the left end points and right end points to over-under evaluate, you can show that
$$1^p+2^p+...+(n-1)^p \leq \int_{1}^n x^p dx \leq 2^p+3^p+...+n^p$$
This is basically the idea behind the integral test.
It is the Riemann sum for $I=\int_0^1x^pdx$
Here is another solution, which uses very few tools from analysis.
Let $P(X)=X^{p+1}-(X-1)^{p+1}$. Expanding with the binomial theorem, you get
$$P(X)=(p+1)X^p +Q(x) \,;\, \deg Q <p$$
Therefore
$$X^p =\frac{1}{p+1} \left(X^{p+1}-(X-1)^{p+1} \right)-\frac{1}{p+1} Q(X)$$
And hence $$\sum_{k=1}^n k^{p}= \frac{1}{p+1} (n^{p+1}-1^{p+1}) -\sum_{k=1}^n\frac{1}{p+1} Q(k)$$
To complete the problem, all you have to prove is that if $\deg q <p$ then $$\lim_n \frac{\sum_{k=1}^n\frac{1}{p+1} Q(k)}{n^{p+1}}=0$$
This is easy because it is enough to observe that if $k \leq p-1$ we have
$$0 \leq \frac{1^k+2^k+...+n^k}{n^{p+1}} \leq \frac{n^k+n^k+...+n^k}{n^{p+1}}=\frac{n^{k+1}}{n^{p+1}}$$
We just need the trivial inequality: $$ p!\binom{n}{p}\leq n^p \leq p!\binom{n+p}{p}\tag{1} $$ and the identity (it can be proved by induction): $$ \sum_{k=p}^{N}\binom{k}{p}=\binom{N+1}{p+1},\tag{2} $$ from which: $$\frac{p!}{n^{p+1}}\binom{n+1}{p+1}\leq \frac{1}{n^{p+1}}\sum_{k=1}^{n}k^p \leq \frac{p!}{n^{p+1}}\binom{n+p+1}{p+1}. \tag{3}$$ Now the claim follows from squeezing.
You can use Faulhaber's Formula which states: $$ \sum^n_{k=1}{k^p}=\frac{1}{p+1}\sum^p_{j=0}{(-1)^j{p+1 \choose j}B_j\ n^{p+1-j}} $$ Where $B_j$ is the $j$th Bernoulli number. Dividing through by $n^{p+1}$, you get that asymptotically the only term that matters is when $j=0$ where the value is $1$ otherwise the sum converges to $0$. Overall the sum is therefore: $$ \lim_{n\to \infty} \frac{1^p + 2^p + \ldots + n^p}{n^{p+1}} = \frac{1}{p+1} $$
