Limit of the sequence $\frac{1^k+2^k+...+n^k}{n^{k+1}}$ How would someone find the limit of the sequence $a_n = \frac{1^k+2^k+...+n^k}{n^{k+1}}, k \in \mathbb{N}$ as $n$ goes to Infinity?
Can someone give me maybe a hint where to start?
 A: By definition of the Riemann integral we have:
$$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n \left(\frac{i}{n}\right)^k=\int_0^1 x^k dx=\frac{1}{k+1}$$
A: We will prove that $\lim_{n\to\infty} a_n = \frac{1}{k+1}$ by elementary means by showing that $$1^k + 2^k + \ldots + n^k$$
can be written as a polynomial of degree $k+1$ in $n$ with leading term $\frac{n^{k+1}}{k+1}$.
Define the following polynomial (of degree $k$ in $n$)
$$p_k(n) = {n\choose k}k! = n(n-1)(n-2)\ldots(n-k+1) = n^k + a_{k-1,n}n^{k-1} + \ldots + a_{1,n}n + a_{0,n}$$
We can now choose $b_{i}$ s.t.
$$\sum_{i=0}^{k}b_{i}p_i(n) = n^k$$
This is done by taking $b_{k}=1$ and then choosing $b_{i}$ (from $i=k-1$ down to $i=0$) s.t. the coefficient of the $n^i$ term vanishes.
We can now write
$$ 1^k + 2^k+\ldots + n^k = \sum_{m=0}^n\sum_{i=0}^kb_{i}p_i(m)$$
and by using the identity $\sum_{m=0}^n{m\choose i}={n+1\choose i+1}$ we finally arrive at
$$ 1^k + 2^k+\ldots + n^k = \sum_{i=0}^kb_{i}{n+1\choose i+1}i! = \sum_{i=0}^k\frac{b_{i}p_{i+1}(n+1)}{i+1} = \frac{n^{k+1}}{k+1} + A_kn^k + \ldots + A_1n$$
and the claim follows. I might add that this proof can with some more work be made constructive by deriving the explicit form for all the $A_{i}$ coefficients and thereby giving us an explcit formula for $\sum_{i=1}^n i^k$.
A: We have by the Riemann sum
$$\sum_{i=1}^n i^k=n^{k+1}\frac1n\sum_{i=1}^n \left(\frac in\right)^k\sim_\infty n^{k+1}\int_0^1 x^kdx=\frac{n^{k+1}}{k+1}$$
so we see easily that
$$\lim_{n\to\infty}a_n=\frac1{k+1}$$
