Q: Series with floor function - convergent? I am trying to figure out where does the following series converge to as $n$ goes to infinity (if it doest at all)
$$\frac{1}{n} \sum^{n}_{t=\lfloor \rho n \rfloor +1} \frac{n}{t}$$
where $\rho$ is from $(0,1)$.
I tried sandwiching it, but didn't really work. How else can I attack this problem? 
Thanks in advance!

A: The $\frac{1}{n}$ and the $n$ in the numerator cancel each other out, don't they?
All you are left with is $H_{n} - H_{k-1}$, $k$ being the lower bound.
$H_n$ is the harmonic number $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \approx \log n + \gamma$. 
Thus your sum is asymptotically $\log \dfrac{n}{k}$, and so will converge to $-\log \rho$ for any $\rho \in (0,1)$.

A: There is no reason to write a sum in such a way, unless you want to make it clear that it is a Riemann sum associated with 
$$ \int_{\rho}^{1}\frac{dx}{x} = \color{red}{-\log\rho}.$$
