# Showing convergence

1. Jun 9, 2010

### vipertongn

1. The problem statement, all variables and given/known data
(ln(n))^2/sqrt(n)

2. Relevant equations
Various series tests

3. The attempt at a solution
The answer showed that the series converges to 0, I know I can rule out the nth term test because the limit is 0. However, I'm not sure how to go about this since it's a logarithmic problem. Can someone direct me as to which test to use to get to thsi? Maybe comparison?

2. Jun 9, 2010

### estro

*Good idea to use latex: $$\sum_{n=1}^\infty \frac{\ln^2{n}} {\sqrt{n}}$$

Last edited: Jun 9, 2010
3. Jun 9, 2010

### lanedance

is this a series or a sequence?

4. Jun 9, 2010

### vipertongn

its a sequence

5. Jun 9, 2010

### vipertongn

what exactly should i compare it to, (im not good with natural logs) I was thinking maybe 1/sqrt(n) but that doesn't show it convergest to 0

6. Jun 9, 2010

### estro

1. What the comparison test says?

2. What is the relation between $$\frac {1} { \sqrt{x} }\ \ and\ \ \frac{\ln^2{x}}{\sqrt{x}}\ ?$$

7. Jun 9, 2010

### lanedance

if the sequence is:
$a_n = \frac{ln(n)^2}{\sqrt{n}}$

how about seeing whether you could show, for some n>N that:
$ln(n)<n^{1/4}$

Last edited: Jun 9, 2010
8. Jun 9, 2010

### vipertongn

wait...OHHH to prove convergence if its a series, u use a series test, but this is just a sequence!!! that means i just need to take the limit to infinity and if its a finite number it should converge!!! :D am i right?

9. Jun 9, 2010

### estro

You converge to a solution...=)
Look carefully what the comparison test means and says.

10. Jun 9, 2010

### vipertongn

if i recall, comparison test is when you compare a sequence that you don't know to a sequence that u do know converges/diverges

11. Jun 9, 2010

### estro

Last edited: Jun 9, 2010
12. Jun 9, 2010

### lanedance

that stuff is for series, so whether the sum converges, say
$S_n = \sum_n a+n = \sum_n \frac{ln(n)^2}{\sqrt{n}}$

if this is actually just a sequence of a_n, then showing a limit exists is sufficient
$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{ln(n)^2}{\sqrt{n}}$

vipertongn - it think you should make sure which it is... in this case the sequence converges to zero, however the series diverges

13. Jun 9, 2010

### vipertongn

it is a sequence, what confused me whas the fact that i find the series to diverge badly but the solutoins said it converged to zero :3 I understand it now

14. Jun 9, 2010

### estro

$$\frac {1} { \sqrt{n} }<\frac{\ln^2{n}}{\sqrt{n}},\ \forall\ n>e$$

$$\sum_{n=e}^\infty \frac{1}{\sqrt{n}}\ is\ divergent\ so\ from\ the\ comparison\ test\ \sum_{n=e}^\infty \frac{\ln^2{n}}{\sqrt{n}}\ is\ also\ divergent\$$

Both sequences converge to 0, but don't confuse the convergence of sequences to convergence of their series.
It seems to me you're not familiar enough with the theory.

Last edited: Jun 9, 2010