Infinite series.

[sylvos]

I'm not sure exactly where my problem is, it may be misusing l'hopital's rule. anyway the Series is:
The sum from n = 1 -> infinity of 1/sqrt(3n-2) does it diverge or converge?
So I try and use limit comparison to the divergent series 1/sqrt(n) (p-series with p=1/2<1)
But here's where my problem starts. dividing by 1/sqrt(n) gives sqrt(n)/sqrt(3n-2) and the limit at infinity is infinity/infinity isn't? so I try lhopitals rule but gets to (1/2)(n^-1/2)/(1/2)((3n-2)^-1/2)(3) but doesn't this just lead me to another indeterminate limit with no way out? Where is my error?

 

[soroban]

Hello, sylvos!

\(\displaystyle S \;=\;\sum^{\infty}_{n=1} \frac{1}{\sqrt{3n-2}}\)

\(\displaystyle \text{Note that: }\:\frac{1}{\sqrt{3x-2}} \;>\;\frac{1}{\sqrt{3n}} \;=\;\frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{n}}\)

\(\displaystyle \text{Hence: }\:S \;>\;\frac{1}{\sqrt{3}}\sum\frac{1}{n^{\frac{1}{2}}} \;\;\text{ which is a divergent }p\text{-series}\)

 

[sylvos]

Ok
So unless I made an error that the improper integral is equal to the limit as b->infinity of 6sqrt(3b-2) - 6sqrt(3(1)-2)=infinity so the series diverges yes?

And soroban that is utilizing direct comparison correct? The thing is this problem was in the limit comparison test section of my textbook though, and I was wondering how to use the test properly. Can it be applied to this series ?

 

[daon]

Here
\(\displaystyle \frac{\sqrt{n}}{\sqrt{3n-2}} = \sqrt{\frac{n}{3n-2}} = \sqrt{\frac{1}{3-\frac{2}{n}}} \to ???\)

 

[sylvos]

Ah ok excellent, thanks a a lot daon.