Limit $\lim_{n\to \infty} \frac{1}{n^{n+1}}\sum_{k=1}^n k^p$ The evaluation of $$\lim_{n\to \infty} \frac{1}{n^{n+1}}\sum_{k=1}^n k^p$$
as an integral ( evaluating this as a right Riemann sum ) requires the form $f(a + k\cdot \Delta x_i)$, where $a$ is our lower bound of integration, and $\Delta x_i$ is the length of our subinterval, on some partition of the interval of integration. However, the given expression does not seem to give way to any meaningful expression for this. I think it's safe to assume our lower bound is $0$ though. 
The bigger problem is the interpretation of $\frac{1}{n^{n+1}}$. I tried $$\frac{1}{n^{n+1}}=\frac{1}{e^{\ln(n^{n+1})}}=\frac{1}{e^{(n+1)\cdot \ln(n)}}$$
and since the inverse of $e^{x}$ is $\ln(x)$, we'd have $$\ln((n+1) \cdot \ln(n)=\ln(n+1)+\ln(\ln(n))$$
which doesn't seem very useful either. I'd appreciate any advice.
 A: The reason why your approach seems to fail is because this expression is trivially just $0$. I'm assuming there's been a transcription error, but if not, then we have $$\begin{align}&\lim_{n\to\infty}\frac{1}{n^{n+1}}\sum_{k=1}^nk^p\\
=&\lim_{n\to\infty}\frac1{n^{n-p}}\left(\frac 1n\sum_{k=1}^n\left(\frac kn\right)^p\right)\end{align}$$
where the right bracket is the integral expression you're looking for (which is bounded by $1$), and the left goes to $0$.
Perhaps your sum should be $$\lim_{n\to\infty}\frac{1}{n^{p+1}}\sum_{k=1}^nk^p$$ in which case you just want the right bracket, which is evaluatable as a Riemann Sum.
A: Hint: $\int_1^{n} x^{p}dx= \sum\limits_{k=2}^{n}\int_{k-1}^{k}x^{p}dx$. Note that $\int_{k-1}^{k}x^{p}dx$ lies between $k^{p}$ and $(k-1)^{p}$. Using this, conclude that $\sum\limits_{k=1}^{n} k^{p}$ lies between $\int_1^{n} x^{p}dx+n^{p}$ and $\int_1^{n} x^{p}dx+1 -(n+1)^{p}$. Of course $\int_1^{n} x^{p}dx=\frac {n^{p+1}-1} {p+1}$. This should make it easy for you to find the limit for different values of $p$. 
