- #1
DKnight768
- 4
- 0
Hi. I have been banging my head against this problem for a while and I just don't get it. Maybe (probably) it's something wrong with my logarithm-fu or limit-fu. I just registered to ask this because I couldn't find an answer anywhere, and I've been reading these forums for a while for other problems and they've been a big help.
Use the ratio test to find if [tex]\sum\frac{1}{log(n)}[/tex] converges.
[tex]\sum\frac{1}{log(n)}[/tex]
Well, I tried [tex]\frac{a_{n+1}}{a_{n}}[/tex] and got [tex]\frac{log(n)}{log(n+1)}[/tex].
I can also try [tex]a_{n}=a_{n-1}[/tex] and then it's [tex]\frac{log(n-1)}{log(n)}[/tex], isn't it?
Maybe this is stupid, but can I remove the log() signs from the fraction? I can't, right?
I guess my problem is that I don't know how to compute the limit of [tex]\frac{log(n-1)}{log(n)}[/tex] as n becomes very large. log(n-1) is less than log(n), so maybe it's 0, and since the condition is that c is 0<c<1 it does not converge? I'm not confident at all in that answer. Please help
PS: I hope the latex symbols show up allright...
Homework Statement
Use the ratio test to find if [tex]\sum\frac{1}{log(n)}[/tex] converges.
Homework Equations
[tex]\sum\frac{1}{log(n)}[/tex]
The Attempt at a Solution
Well, I tried [tex]\frac{a_{n+1}}{a_{n}}[/tex] and got [tex]\frac{log(n)}{log(n+1)}[/tex].
I can also try [tex]a_{n}=a_{n-1}[/tex] and then it's [tex]\frac{log(n-1)}{log(n)}[/tex], isn't it?
Maybe this is stupid, but can I remove the log() signs from the fraction? I can't, right?
I guess my problem is that I don't know how to compute the limit of [tex]\frac{log(n-1)}{log(n)}[/tex] as n becomes very large. log(n-1) is less than log(n), so maybe it's 0, and since the condition is that c is 0<c<1 it does not converge? I'm not confident at all in that answer. Please help
PS: I hope the latex symbols show up allright...