Convergence of (ln(n))^2/sqrt(n) - Series Tests Explained

[vipertongn]

Homework Statement

(ln(n))^2/sqrt(n)

Homework Equations

Various series tests

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?

 

[estro]

Think about comparison test.

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

 

[lanedance]

is this a series or a sequence?

 

[vipertongn]

its a sequence

 

[vipertongn]

Think about comparison test.

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

 

[estro]

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

1. What the comparison test says?

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

 

[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}

 

[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?

 

[estro]

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?

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

 

[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

 

[estro]

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

You must be familiar and understand the theory well before moving on...
http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx

 

[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

 

[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

 

[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.