Solving $\lim_{n→∞} \left( \frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1} \right)$ where $k∈\mathbb{N}$ I had to find the limit of the sequence
$$ a_n =\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}$$, where $ k $ is a natural number.
After applying the Stolz theorem, I was able to get here.
$$\lim \:_{n\to \:\infty \:}\left(\frac{\left(k+1\right)\left(n+1\right)^k-\left(\left(n+1\right)^{k+1}-n^{k+1}\right)}{\left(k+1\right)\left(\left(n+1\right)^k-n^k\right)}\right)$$
I tried different ways on continuing from here but unfortunately I am unable to get anywhere. I appreciate any kind of help.
Edit:
For users who may want a solution without the Big O notation, I was able to solve this limit by using the binomial theorem to calculate largest coefficient, since the numerator and denominator are polynomials in $n$, as a user below suggested.
 A: We have
$$a_n =\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}=a_n =\frac{(k+1)(1^k+2^k+\cdot \cdot \cdot +n^k)-n^{k+1}}{(k+1)n^k}$$
and then by Stolz-Cesaro
$$\frac{(k+1)(n+1)^{k}-(n+1)^{k+1}+n^{k+1}}{(k+1)(n+1)^k-(k+1)n^k}$$
and since by binomial expansion

*

*$(k+1)(n+1)^{k}=(k+1)n^k+k(k+1)n^{k-1}+O(n^{k-2})$

*$(n+1)^{k+1}= n^{k+1}+(k+1)n^k+\frac{k(k+1)}2n^{k-1}+O(n^{k-2})$
we have
$$\frac{(k+1)(n+1)^{k}-(n+1)^{k+1}+n^{k+1}}{(k+1)(n+1)^k-(k+1)n^k} = \frac{k(k+1)n^{k-1}-\frac{k(k+1)}2n^{k-1}+O(n^{k-2})}{k(k+1)n^{k-1}+O(n^{k-2})}=$$
$$\frac{\frac12+O(1/n)}{1+O(1/n)}\to \frac12$$

Edit
As an alternative, directly by Faulhaber's formula since
$$1^k+2^k+\cdot \cdot \cdot +n^k = \frac{n^{k+1}}{k+1}+\frac12 n^k+O(n^{k-1})$$
we have
$$\frac{1^k+2^k+\cdot \cdot \cdot +n^k}{n^k}-\frac{n}{k+1}=\frac{n}{k+1}+\frac12+O(1/n) -\frac{n}{k+1}=\frac12+O(1/n) \to \frac12$$
A: The numerator and denominator are polynomials in $n$.
using the binomial theorem the largest coefficents can be calculated. They are
$$\frac{\frac{k(k+1)}{2}n^{k-1}+\cdots}{k(k+1)n^{k-1}+\cdots }\to \frac{1}{2}$$
A: There is another way to show more than the limit itself if you are familiar with generalized harmonic numbers since
$$\sum_{i=1}^n i^k=H_n^{(-k)}$$ and for large values of $n$
$$H_n^{(-k)}=n^k \left(\frac{n}{k+1}+\frac{1}{2}+\frac{k}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-k)$$ which makes
$$a_n=\left(\frac{1}{2}+\frac{k}{12
   n}+O\left(\frac{1}{n^3}\right)\right)+\frac{\zeta (-k) }{n^{k} }\sim \frac{1}{2}+\frac{k}{12
   n}$$
