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

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.
