$\displaystyle \sum_{n=1}^{\propto } \frac{3\sqrt{n}}{n}$
How to prove this series diverges?
Many thanks!
$\displaystyle \sum_{n=1}^\infty\frac{3\sqrt{n}}{n}=$$\displaystyle \sum_{n=1}^\infty\frac{3}{\sqrt{n}}$
$\displaystyle \sum_{n=1}^\infty\frac{1}{\sqrt{n}}$ is a p-series with $\displaystyle p=\frac{1}{2}$. So, $\displaystyle \sum_{n=1}^\infty\frac{1}{\sqrt{n}}$ diverges and consequently, $\displaystyle \sum_{n=1}^\infty\frac{3}{\sqrt{n}}$ diverges.
A 'direct' proof that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges is very comfortable. If we consider the partial sum of the first k terms we note that...
$\displaystyle S_{k}=\sum_{n=1}^{k} \frac{1}{\sqrt{n}} \ge k\ \frac{1}{\sqrt{k}}= \sqrt{k}$ (1)
... so that $\displaystyle \lim_{k \rightarrow \infty} S_{k}= \infty$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$