Sequence/series convergence

Question

I have the bellow sequence with n from $\Bbb N$ and I need to show if it converges and if is a Cauchy sequence.

$$a_n = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}$$

Can't I turn it into a series like the one bellow, apply the p-test and see that it diverges?

$$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$

Also for the sequences of the same form, that don't diverges, can't I turn into a series, apply some test, see if are converging, calculate the sum which will be the $a_n$ limit?

To show if is Cauchy I just need to show that are converging or diverging since all converging are Cauchy sequences.

Edit: My question is not about if this particular sequence converge, it is if I can put this sequence as a series (and other of this form) and apply the series tests, in this case the easiest and faster one would the p-test (if p > 1, p being 1/2 here), if the result will be the same, rather than applying the Cauchy eplison thing.

Answer 1

$$ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\gt \frac{1}{\sqrt n} + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \cdots+ \frac{1}{\sqrt{n}}$$ Hence we have $$ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\gt \sqrt n.$$

Answer 2

You can note that for any natural $n$,

$$\frac{1}{n} \le \frac{1}{\sqrt{n}}$$

and therefore your series must be greater than the harmonic series

$$ \sum_{n=0}^{\infty} \frac{1}{n}$$

but the harmonic series diverges and so does yours. That is because:

$$ \forall\ n \in \mathbb{N}\ a_n \ge b_n \rightarrow \sum_{i=0}^{k} a_n \ge \sum_{i=0}^{k} b_n$$

Taking the limit of $k \rightarrow \infty$ it follows that your series diverges because it is greater than the harmonic series.