If $l_n=\sum_{k=1}^{n}\frac{1}{k}\sim \log n$, find $\inf_n \frac{l_n}{n}.$

Question

Let $l_n=\sum_{k=1}^{n}\frac{1}{k}\sim \log n$. I m stack to find $\displaystyle\inf_n \frac{l_n}{n}.$

Answer 1

The sequence $\frac {l_n} n$ is decreasing so its infimum is its limit; the limit is $0$ by Cesaro's Theorem.

To prove that $\frac 1 n \sum\limits_{k=1}^{n} \frac 1 k \geq \frac 1 {n+1} \sum\limits_{k=1}^{n+1} \frac 1 k $ just rewrite this inequality as $\frac n {n+1} \leq \sum\limits_{k=1}^{n} \frac 1 k$ which is true becasue the left side is at most $1$ and the right side is at least $1$.

Answer 2

If we consider $l_n = log(n)$ then the sequence $\frac{l_n}{n}$ would be {$0,0.1505,0.159,0.1505,0.1398,.....,0.1,...$}. Now as $log(n)$ increases slowly as compared to $n$ and also evident from the sequence that it is converging sequence it will converge to $0$ $\therefore$ the $inf \frac{l_n}{n}$ would be $0$.