Computing a limes via integration Currently, I am studying for an exam (topics: real analysis, integration etc.). I came across the following exercise:
Let $\lambda>-1$ and let $(a_n)$ be the sequence defined by
$$a_n=\frac{1^{\lambda}+2^{\lambda}+...+n^{\lambda}}{n^{\lambda+1}}$$
Compute the limes of $(a_n)$.
For $\lambda=0$, the problem is trivial. If $\lambda>0$ and if we make use of appropriate Riemann sums, it is not hard to see that
$$\lim a_n=\int_0^1 x^{\lambda}dx$$
My question: How to find the limes for the case that $\lambda<0$? 
The problem I have here is that the trick via integration seems not to work in the same fashion as in the case $\lambda>0$ since for $\lambda<0$, the map $x \mapsto x^{\lambda}$ is not defined for $x=0$, so we can not integrate this map on the intervall $[0,1]$. Any suggestions?
 A: If $-1 < \lambda < 0$ then the improper Riemann integral is finite.  
For example,
$$\int_0^1 x^{-1/2} \, dx = \lim_{c \to 0}\int_c^1 x^{-1/2} \, dx = \lim_{c \to 0}2x^{1/2}|_c^1 = 2.$$
This coincides with the limit of the right-hand Cauchy sum:
$$\lim_{n \to \infty}\sum_{k=1}^n \frac{k^{-1/2}}{n^{-1/2 +1}} =\lim_{n \to \infty} \frac1{\sqrt{n}}\sum_{k=1}^n \frac1{\sqrt{k}}=2.$$
Note that
$$\frac{2}{\sqrt{k}+\sqrt{k+1}} \leqslant \frac1{\sqrt{k}} \leqslant \frac{2}{\sqrt{k-1}+\sqrt{k}}   \\ \implies 2(\sqrt{k+1}-\sqrt{k}) \leqslant  \frac1{\sqrt{k}}  \leqslant 2(\sqrt{k}-\sqrt{k-1}) \\ \implies 2(\sqrt{n+1} - 1)\leqslant \sum_{k=1}^n \frac1{\sqrt{k}} \leqslant 2\sqrt{n} \\ \implies 2\left(\sqrt{1+1/n} - 1/\sqrt{n}\right)\leqslant \frac1{\sqrt{n}}\sum_{k=1}^n \frac1{\sqrt{k}} \leqslant 2,$$
and the limit is shown to exist and equal $2$ using the squeeze principle.
More generally, if $f:(0,1] \to \mathbb{R}$ is monotonic and the improper integral over $[0,1]$ is finite, then the sequence of right-hand sums converges to the value of the improper integral.  
