Evaluating $\lim\limits_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$ Evaluate  $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$

The first couple of terms of the Euler-Maclaurin Sum Formula can be derived as
$$
\begin{align}
\sum_{k=1}^nk^p
&=\int_{0^+}^{n^+}x^p\,\mathrm{d}\!\left\lfloor x\right\rfloor\tag1\\
&=\int_{0^+}^{n^+}x^p\,\mathrm{d}(x-\{x\})\tag2\\
&=\tfrac1{p+1}n^{p+1}-\int_{0^+}^{n^+}x^p\,\mathrm{d}\!\left(\{x\}-\tfrac12\right)\tag3\\
&=\tfrac1{p+1}n^{p+1}-\left[x^p\left(\{x\}-\tfrac12\right)\right]_{0^+}^{n^+}+p\int_0^n\left(\{x\}-\tfrac12\right)x^{p-1}\,\mathrm{d}x\tag4\\[6pt]
&=\tfrac1{p+1}n^{p+1}+\tfrac12n^p+O\!\left(n^{p-1}\right)\tag5
\end{align}
$$
Explanation:
$(1)$: write sum as a Riemann-Stieltjes Integral
$(2)$: $\left\lfloor x\right\rfloor=x-\{x\}$
$(3)$: use $\{x\}-\frac12$ because its integral is periodic
$(4)$: Integrate by Parts
$(5)$: see below
Proof of $(5)$:
$$
\begin{align}
\left|\,p\int_0^n\left(\{x\}-\tfrac12\right)x^{p-1}\,\mathrm{d}x\,\right|
&=\left|\,p\int_0^n\left(\{x\}-\tfrac12\right)\left(x^{p-1}-\left\lfloor x\right\rfloor^{p-1}\right)\,\mathrm{d}x\,\right|\tag6\\
&=\left|\,p\sum_{k=0}^{n-1}\int_0^1\left(x-\tfrac12\right)\left((k+x)^{p-1}-k^{p-1}\right)\,\mathrm{d}x\,\right|\tag7\\
&\le p\sum_{k=0}^{n-1}\tfrac12\left((k+1)^{p-1}-k^{p-1}\right)\tag8\\[6pt]
&=\tfrac{p}2n^{p-1}\tag9
\end{align}
$$
Explanation:
$(6)$: the integral of $\{x\}-\tfrac12$ over a unit interval is $0$
$(7)$: for $(k,x)\in\mathbb{Z}\times[0,1)$, $\{k+x\}=x$ and $\lfloor k+x\rfloor=x$
$(8)$: for $(k,x)\in\mathbb{Z}\times[0,1)$, $\left|x-\frac12\right|\le\frac12$ and $\left|(k+x)^{p-1}-k^{p-1}\right|\le(k+1)^{p-1}-k^{p-1}$
$(9)$: Telescoping Sum
Divide $(5)$ by $n^p$, subtract $\frac{n}{p+1}$, and take the limit.
