Test for divergence or convergence.
(-1)(bsubn) ln(n)/(sqrt(n))
I used the alternating series test, and proved that (bsubn+1) is smaller than (bsubn) but the limit as n---infinity of (bsubn) does not equal 0.
My book is saying it does converge though...

Question

Test for divergence or convergence.
(-1)(bsubn) ln(n)/(sqrt(n))
I used the alternating series test, and proved that (bsubn+1) is smaller than (bsubn) but the limit as n--->infinity of (bsubn) does not equal 0.
My book is saying it does converge though...

anonymous · Answer

I was thinking to then try the ratio test, but that doesnt seem to be working either...

psymon · Answer

Well, if your bsub n is ln(n)/√(n), then it limit is 0, which fits the condition needed.

anonymous · Answer

wouldnt that be infinity over inifinity?

psymon · Answer

You need to do l'hopitals rule.

anonymous · Answer

yeah, your right. Ok lemme try that

anonymous · Answer

ok..i got 2/(sqrt(n)) when i did L'hospitals rule...and it does come out to 0 then...

psymon · Answer

Which is what I got : ) And you said you already proved that bsub n+1 was less than b sub n, so sounds like you have your convergence.

anonymous · Answer

thanks psymon!

psymon · Answer

Yep, np :3