1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It diverges, thus is not a Cauchy sequence. $\endgroup$ – Cave Johnson Nov 7 '16 at 15:53
  • 6
    $\begingroup$ Please note that it is Cauchy and not Catchy :) $\endgroup$ – RGS Nov 7 '16 at 15:55
  • 1
    $\begingroup$ @RSerrao Edited $\endgroup$ – imranfat Nov 7 '16 at 16:12
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Did you mean $\le$ in your first equation? $\endgroup$ – Keba Nov 7 '16 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.