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 result is more general.

Fact: For any function $f$ regular enough on $[0,1]$, introduce
  $$
A_n=\sum_{k=1}^nf\left(\frac{k}n\right)\qquad B=\int_0^1f(x)\mathrm dx\qquad C=f(1)-f(0)
$$
  Then,
  $$
\lim\limits_{n\to\infty}A_n-nB=\frac12C
$$

For any real number $p\gt0$, if $f(x)=x^p$, one sees that $B=\frac1{p+1}$ and $C=1$, which is the result in the question.
To prove the fact stated above, start from Taylor formula: for every $0\leqslant x\leqslant 1/n$ and $1\leqslant k\leqslant n$,
$$
f(x+(k-1)/n)=f(k/n)-(1-x)f'(k/n)+u_{n,k}(x)/n
$$
where $u_{n,k}(x)\to0$ when $n\to\infty$, uniformly on $k$ and $x$, say $|u_{n,k}(x)|\leqslant v_n$ with $v_n\to0$. Integrating this on $[0,1/n]$ and summing from $k=1$ to $k=n$, one gets
$$
\int_0^1f(x)\mathrm dx=\frac1n\sum_{k=1}^nf\left(\frac{k}n\right)-\frac1n\int_0^{1/n}u\mathrm du\cdot\sum_{k=1}^nf'\left(\frac{k}n\right)+\frac1nu_n
$$
where $|u_n|\leqslant v_n$. Reordering, this says that
$$
A_n=nB+\frac12\frac1n\sum_{k=1}^nf'\left(\frac{k}n\right)-u_n=nB+\frac12\int_0^1f'(x)\mathrm dx+r_n-u_n
$$
with $r_n\to0$, thanks to the Riemann integrability of the function $f'$ on $[0,1]$. The proof is complete since $r_n-u_n\to0$ and the last integral is $f(1)-f(0)=C$.
