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

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.
