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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!

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2 Answers 2

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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 $\rho \in (0,1)$.

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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 $$ \int_{\rho}^{1}\frac{dx}{x} = \color{red}{-\log\rho}.$$

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