Checking the Convergence of the sequence $a_n=\frac{1}{n}\left(1+\frac{1}{2}+\frac{1}{3}+....\frac{1}{n}\right)$

Question

Check the Convergence of the sequence $a_n=\frac{1}{n}\left(1+\frac{1}{2}+\frac{1}{3}+....\frac{1}{n}\right)$

My effort:

I actually took the help of inequality(not well known I guess) $$H_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}<2\sqrt{n}, \forall n \in \mathbb{N}$$

Now we have $a_n \geq 0$ and $a_n=\frac{H_n}{n} < \frac{2\sqrt{n}}{n}=\frac{2}{\sqrt{n}}$

Thus we have $$0 \leq a_n <\frac{2}{\sqrt{n}}$$ By Sandwich Theorem, we end up with $$\lim a_n=0$$

But is there a way without Sandwich Theorem?

Answer 1

As very simple bounds, you could use $$\displaystyle \log(n) + \frac{1}{n} \;\lt \; H_n \; \lt \; \log(n) + 1$$