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  $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$
 A: another method, using Stolz–Cesàro theorem:
let ${ x }_{ n }=\left( p+1 \right) \left( { 1 }^{ p }+{ 2 }^{ p }+...+{ n }^{ p } \right) -{ n }^{ p+1 },{ y }_{ n }=\left( p+1 \right) { n }^{ p }$
$$\lim _{ x\rightarrow \infty  }{ \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } = } \lim _{ x\rightarrow \infty  }{ \frac { \left( p+1 \right) { \left( n+1 \right)  }^{ p }-{ \left( n+1 \right)  }^{ p+1 }+{ n }^{ p+1 } }{ \left( p+1 \right) \left( { \left( n+1 \right)  }^{ p }-{ n }^{ p } \right)  } = } \\ =\lim _{ x\rightarrow \infty  }{ \left( \frac { \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right)  }{ 2 } { n }^{ p-2 }+...+1 \right)  }{ \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right)  }{ 2 } { n }^{ p-2 }+...+1-{ n }^{ p } \right)  }  \right) + } \\ +\frac { -{ n }^{ p+1 }-\left( p+1 \right) { n }^{ p }-\frac { p\left( p+1 \right)  }{ 2 } { n }^{ p-1 }-...-1+{ n }^{ p+1 } }{ \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right)  }{ 2 } { n }^{ p-2 }+...+1-{ n }^{ p } \right)  } $$
let's cobmine all coefficients of n,then divide numerator and denominator by $n^{ p-1 }$ and define sum of the  all terms  no more -1 power with $o\left( \frac { 1 }{ n }  \right) $
$$\\ \\ \lim _{ x\rightarrow \infty  }{ \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } = } \lim _{ x\rightarrow \infty  }{ \frac { \frac { p\left( p+1 \right)  }{ 2 } +o\left( \frac { 1 }{ n }  \right)  }{ p\left( p+1 \right) +\left( \frac { 1 }{ n }  \right)  } =\frac { 1 }{ 2 }  } \\ $$
A: 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 $f(1)-f(0)=C$.
A: 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.
A: This is a nice little question. I am assuming that $p \in \mathbb{Z}^+$, though same could be said about it when $p \notin \mathbb{Z}^+$. Before getting to the answer lets experiment a bit for small positive integers $p$. To start off, you could try for some values $p$.
For $p=1$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n(n+1)}{2}}{n} - \frac{n}{1+1} \right) = \lim_{n \rightarrow \infty} \frac12 = \frac12$$
For $p=2$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n(n+1)(2n+1)}{6}}{n^2} - \frac{n}{2+1} \right) = \lim_{n \rightarrow \infty} \left(\frac{(n+1)(n+1/2)}{3n} - \frac{n}3 \right)\\ = \lim_{n \rightarrow \infty} \left(\frac{n}3 + \frac12 + \frac1{6n} - \frac{n}3  \right)= \frac12$$
For $p=3$, we get $$\lim_{n \rightarrow \infty} \left(\frac{ \dfrac{n^2(n+1)^2}{4}}{n^3} - \frac{n}{3+1} \right) = \lim_{n \rightarrow \infty} \left(\frac{n^2 + 2n + 1}{4n} - \frac{n}4 \right)\\ = \lim_{n \rightarrow \infty} \left(\frac{n}4 + \frac12 + \frac1{4n} - \frac{n}4  \right)= \frac12$$
Hence, we would guess that it is $\dfrac12$ independent of $p$. And this turns out to be right.
Let us denote $1^p + 2^p + \cdots n^p = P_p(n)$. This is a polynomial of degree $p+1$ and is given by
$$P_p(n) = \frac1{p+1} \sum_{k=0}^p \dbinom{p+1}{k} B_k n^{p+1-k}$$ where $B_k$ are the Bernoulli numbers. These polynomials are related to the Bernoulli polynomials and there are some really nice results on these polynomials and more can be found here.
Hence, $$\dfrac{P_p(n)}{n^{p}} = \dfrac1{p+1} \sum_{k=0}^p \dbinom{p+1}{k} B_k n^{1-k} = \dfrac1{p+1} \left(B_0 n + (p+1) B_1 + \mathcal{O} \left(\frac1n\right) \right)$$
where $B_0 = 1$ and $B_1 = \frac12$.
What you are looking for is $$\lim_{n \rightarrow \infty} \left(\dfrac{P_p(n)}{n^{p}} - \dfrac{n}{p+1} \right) = \lim_{n \rightarrow \infty} \left(\dfrac1{p+1} \left(n + (p+1) B_1 + \mathcal{O} \left(\frac1n\right) \right) - \dfrac{n}{p+1} \right)\\ = \lim_{n \rightarrow \infty} \left(B_1 + \mathcal{O} \left(\dfrac1n \right)\right)= B_1 = \frac12$$ independent of $p$.
Users Did and Ragib Zaman have provided excellent solutions. You might also want to look at Euler–Maclaurin formula which is of significance in this context.
A: This is just to show an alternative  proof of the fact stated by @Did as I was not able to follow his argument completely (not for lack of clarity on Did's part but of my lack of mental capacity I suppose).
Lebesgue integration by parts yields
$$
\sum_{a<k\leq b}g(k)=\int^b_a g(x)\,dx +\int^b_a g'(x)\{x\}\,dx+g(a)\{a\}-g(b)\{b\}
$$
for any absolutely continuous function $g$ on $[a,b]$, where $\{x\}=x-\lfloor x\rfloor$ is the fractional part function.
Letting $g(x)=f(x/n)$ for $0\leq x\leq n$ we obtain
\begin{align}
\sum_{0<k\leq n}f(k/n)&=\int^n_0 f(x/n)\,dx + \frac1n\int^n_0f'(x/n)\{x\}\,dx\\
&=n\int^1_0f(x)\,dx +\int^1_0 f'(x)\{nx\}\,dx
\end{align}
Now, the function $x\mapsto\{x\}$ is 1-periodic. An application of Fejer's lemma yields
$$\lim_{n\rightarrow\infty}\int^1_0 f'(x)\{nx\}\,dx=\Big(\int^1_0\{x\}\,dx\Big)\int^1_0f'(x)\,dx=\frac{f(1)-f(0)}{2}$$
Hence
$$\sum^n_{k=1}f(k/n) - n\int^1_0f(x)\,dx\xrightarrow{n\rightarrow\infty}\frac{f(1)-f(0)}{2}$$
A: If we draw the graph of $x^p$ from $x=1$ to $x=n,$ divide it into unit length intervals and approximate each segment of area by a trapezium (this is known as the trapezoidal rule) then we see that $$\int^n_1 x^p dx \approx \sum_{k=1}^n k^p - \frac{n^p+1}{2}.$$ The integral on the left is precisely $\displaystyle \frac{n^{p+1} -1}{p+1},$ so for large $n$ (where the major contribution is from the dominant terms) we have $$\sum_{k=1}^n k^p \approx \frac{n^{p+1}}{p+1} + \frac{n^p}{2}$$ so your limit is $1/2.$

For a precise solution, we need the error term along with the trapezoidal rule, which is derived here. It gives : $$\int^b_a f(x) dx = \frac{b-a}{2} ( f(a) + f(b) ) - \frac{(b-a)^3 }{12} f''(\zeta) $$ for some $\zeta \in [a,b].$ For $f(x)=x^p$ we have $f''(x) = p (p-1)x^{p-2}$ which is largest at $x=b$, the right end point. So the sum of the error terms in our application of the trapezoidal rule is at largest $$\frac{p(p-1)}{12} (2^{p-2} + 3^{p-2} + \cdots + n^{p-2}).$$ The sum in the brackets is overestimated by $\int^{n+1}_1 x^{p-2} dx= \frac{(n+1)^{p-1}-1}{p-1},$ so we get that $$\sum_{k=1}^n k^p = \frac{n^{p+1}}{p+1} + \frac{n^p}{2} + E_n$$ where $E_n$ is an error term that satisfies $\displaystyle \lim_{n\to\infty} \frac{E_n}{n^p} = 0$ which proves your limit.
A: Hint for another method:
You can use Faulhaber's formula:
$$1^p+2^p+\dots+n^p=\frac1{p+1}\sum_{k=0}^p\binom{p+1}kB_k\, n^{p+1-k},$$
where $B_k$ denotes the $k$-th Bernoulli number (with the convention that $B_1=-\frac12$).
