Determine if $a_n=\frac{\ln^2(n)}{n^{1/2}}$ is convergent by using the properties of limits

Question

Determine if $\lim\limits_{n \to \infty} a_n$=$\frac{\ln^2(n)}{\sqrt{n}}$ is convergent by using the properties of limits

So I will take the limits of both top and bottom: $\lim\limits_{n \to \infty}\ln^2(n))=\infty$ and $\lim\limits_{n \to \infty} \sqrt{n}= \infty$.

So limit of the fraction should be $1$ but the calculator does not agree with this.

Reattempt:

Derivative of $ln^2(n)$ is $\frac{2\ln(n)}{n}$

Derivative of $\sqrt{n}$ is $\frac{1}{2\sqrt{n}}$

Dividing top and bottom gives $\frac{4}{n}$ so limit is 0.

First your limit is in indeterminate form. Second $\infty/\infty \not=1$ — Kushal Bhuyan, Jun 10 '16 at 12:29
First it can lead to wrong results if students forget the hypotheses (in particular, the denominator must not be $0$ in a neighbourhood of $x_0$, except at $x_0$ itself). Second, when it works, Taylor's formula at order $1$ works. Third, using equivalents is often more efficient. — Bernard, Jun 10 '16 at 14:33

score 3 · Answer 1 · answered Jun 10 '16 at 12:39

3

HINT:

For any number $\alpha>0$, we have

$$\log(x) \le \frac{x^{\alpha}}{\alpha}$$

answered Jun 10 '16 at 12:39

Mark Viola

Super cool. This inequality (for $0 < \alpha < 1/4$) actually provides that the limit vanishes, despite that convergence is relatively slow. – Steven Harding Jun 10 '16 at 13:11
@StevenHarding Thank you! And that is correct. – Mark Viola Jun 10 '16 at 13:25

score 3 · Answer 2 · answered Jun 10 '16 at 12:39

3

It can be easy if you exchange variable n to $ e^{2x} $. Then you can use L'Hospital Rule or just use definition.

answered Jun 10 '16 at 12:39

user346926

I did it but without a substitution. – stackdsewew Jun 10 '16 at 12:42
It's all right, because calculate this derivation is easy. – user346926 Jun 10 '16 at 12:47

2 Answers2