question 2

[rohiiit]

how do u start on this ?

[idgit]

I'm not really sure about i) either.
I know we either have to
a) solve the integral then show it converges (ie has a limit) or diverges (limit is infinity or D.N.E)
or
b) find something less than it that is divergent and therefore it is divergent
or
c) find something greater than it that is convergent and therefore it is convergent

So in trying the first method I get something like x/((ln x)^2) + 2Integral dx/((ln x)^3)    Which doesn't really help me, although the first part seems to be divergent I don't think you can therefore say the whole thing is divergent without knowing about the second interval part.

So then we need to do either b) or c).   So what is larger and convergent or smaller and divergent that can be solved??  Larger would be integral dx/ln x but this ends up having integral dx/((ln x)^2) in it, which is what we want to find in the first place.   Smaller would be dx/((ln x)^3) but this would probably have integral dx/((ln x)^4) in it...  assuming I'm doing this right.

Anyway, I'm at a loss for what integral to pick to compare it with, can anyone give me some hints?   I'm having the same problem with the last question in 1. as well.

[idgit]

has no one started this one yet??  
or is it way to easy to comment?

I think this assignment is hard, personally...   mind you I've thought that most of them were hard.

[Rogue_Knight]

The question only asks you to show that it either diverges or converges.  My guess would be to use the comparison method to show convergence/divergence.

The functions to use however, I'm not so clear on.

[Majin_Blues]

just use functions similar to the question... i can't really post exact formulas since people will use it and there'll be another case of mass plaguarism (that how u spell it?)

there's tons of possible formulas... here's a hint for the second one: as x-> oo, what's sinx bounded by?

also, for the first: x > lnx... i think... try it out... not sure if it helps, i did it differently...

[Rogue_Knight]

What we need to compare is in the form 1/f(x) right?
So note the upper and lower bounds for f(x).
Then this inequality reverses when you take the reciprical and you get your equation to compare it with.