How find this $\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)$ Find this limit

$$\lim_{n\to\infty}n^2\left(\dfrac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\dfrac{1}{k+1}-\dfrac{1}{2n}\right)\tag{1}$$

I can  only solve this limit

$$I=\lim_{n\to\infty}\left(\dfrac{1^k+2^k+\cdots+n^k}{n^k}-\dfrac{n}{k+1}\right)=\dfrac{1}{2}$$
  proof:
  use Stolz therom:let
  $$a_{n}=(k+1)(1^k+2^k+\cdots+n^k)-n^{k+1},b_{n}=(k+1)n^k$$
  then
  $$I=\lim_{n\to\infty}\dfrac{a_{n}}{b_{n}}=\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\dfrac{1}{k+1}\dfrac{(k+1)(n+1)^k-(n+1)^{k+1}+n^{k+1}}{(n+1)^k-n^k}$$
  so
  $$I=\lim_{n\to\infty}\dfrac{1}{k+1}\dfrac{[k(k+1)/2]n^{k-1}+\cdots+k}{kn^{k-1}+\cdots+1}=\dfrac{1}{2}$$
  But for (1),I can't,Thanks.


We know that $\sum_{i=1}^n i^k=s_k(n)$ for some polynomial $s_k(x)=a_0+a_1x+\ldots +a_{k+1}x^{k+1}$. (We know this because the map $\Delta\colon f\mapsto f(X)-f(X-1)$ is a linear map from the $\mathbb Q$-vector space of polynomials of degree $\le k+1$ to the vector space of polynomials of degree $\le k$ and the kernel consists precisely of the onedimensional space of constant polynomials, hence $\Delta$ is onto).
We want to take the limit of 
$$ \begin{align}&n^2\left(\frac{p_k(n)}{n^{k+1}}-\frac{1}{k+1}-\frac1{2n}\right)\\=&n^2\left[\left(a_{k+1}-\frac1{k+1}\right)+\left(a_k-\frac12\right)\frac1n+a_{k-1}\frac1{n^2}+O\left(\frac1{n^3}\right)\right]\\
\end{align}$$
and observe that the result is $a_{k-1}$ (provided that we can confirm that  $a_{k+1}=\frac1{k+1}$ and $a_k=\frac12$).
One can determine the highest $a_j$ step by step by comparing coefficients:
$$\begin{align}n^k &=s_k(n)-s_k(n-1)\\
&=a_{k+1}(n^{k+1}-(n-1)^{k+1})+a_k(n^k-(n-1)^k)+a_{k-1}(n^{k-1}-(n-1)^{k-1})+\ldots \\
&=\hphantom{+\;}a_{k+1}\left((k+1)n^k-\tfrac{(k+1)k}{2}n^{k-1}+\tfrac{(k+1)k(k-1)}{6}n^{k-2}+O(n^{k-3})\right)\\
&\quad+\;a_{k}\left(kn^{k-1}-\tfrac{k(k-1)}{2}n^{k-2}+O(n^{k-3})\right)\\
&\quad+\;a_{k-1}\left((k-1)n^{k-2}+O(n^{k-3})\right)\\
&\quad +\;O(n^{k-3})\\
&=\hphantom{+\;}a_{k+1}(k+1)\cdot n^k\\
&\quad+\;\left(a_kk-a_{k+1}\tfrac{(k+1)k}{2}\right)\cdot  n^{k-1}\\&\quad+\;\left(a_{k-1}(k-1)-a_k\tfrac{k(k-1)}{2}+a_{k+1}\tfrac{(k+1)k(k-1)}{6}\right)\cdot n^{k-2}\\&\quad +\;O(n^{k-3}),\end{align}$$
that is
$$\begin{align}a_{k+1}&=\frac1{k+1}\\
a_k&=\frac1k\cdot a_{k+1}\frac{(k+1)k}{2}=\frac12\\
a_{k-1}&=\frac1{k-1}\left(a_k\frac{k(k-1)}{2}-a_{k+1}\frac{(k+1)k(k-1)}{6}\right)=\frac k{12}.\end{align} $$
Thus the final result is 
$$ \lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\ldots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac1{2n}\right)=\frac k{12}.$$
