What function should I use for comparison

[armolinasf]

Homework Statement

If i have the series ln(k)/k^2 would I compare it to 1/k^2? The reason why I'm confused is that ln(k)/k^2 > 1/k^2 so if its greater than a function that converges it doesn't tell us anything right? So I'm not sure what exactly to compare it with on account of the ln(k) in the numerator. thanks for the help.

 

[eczeno]

hello,

you are right that 1/k^2 will not help us in this case.

the first thing you need to decide is whether you think your series converges or diverges; that will determine the kind of series you want to compare it to. what do you think?

 

[armolinasf]

my guess would be that it converges

 

[eczeno]

ok, then we need a series that converges, but that is bigger than ours.

do you have to use the comparison test, or can yo use any method?

 

[eczeno]

what series have you tried?

 

[blinktx411]

Here's a hint: k^(1/2) > ln(k) for all k.

 

[armolinasf]

Alright if k^1/2 is greater than ln k i could compare it with K^(1/2)/k^2 = 1/k^3/2 which converges since p>1.

One more question: what if it were 1/ln(k)?

 

[eczeno]

good!

well, for 1/ln(k) we again first need to decide if we think it converges or diverges. this one should be pretty easy if we think carefully. if we think it diverges (and hopefully we do ), we need f(k) > ln(k) and 1/f(k) divergent. that shouldn't be too hard to find.

cheers