convergent/divergent with natural log


by dlevanchuk
Tags: natural
dlevanchuk
dlevanchuk is offline
#1
Oct11-09, 03:38 PM
P: 29
Preparing for the math test, and cannot understand what to do for one of my practice problems:

I need to find if the series is convergent or divergent, using test for divergence (divergent if limit doesn't equal 0)

sigma from 1 to infinity of ln((n^2+1)/(2n^2+1))

I see that I should use l'Hopital rule, and show that as n goes to infinity, the limit will equal to 1/2 (and series is diverges), but the ln (ie, natural log) gives me a doubt about my answer...

Also, i figured that I can use one of the natural log properties and rewrite as sigma ln(n^2+1) - sigma ln(2n^2+1), but still i have no idea what to do with ln...

help! what kind of influence does the natural log have on this problem??
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dlevanchuk
dlevanchuk is offline
#2
Oct11-09, 03:40 PM
P: 29
Another idea is that to do a l'Hopital rule on ln((n^2+1)/(2n^2+1)), which will flip my fraction, getting rid of natural log, and do l'Hopital rule 2 times more, and that will make my limit equals to 2, and show that the series is divergent...

Is my logic solid, or does it have a hole somewhere?
daniel_i_l
daniel_i_l is offline
#3
Oct11-09, 04:08 PM
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There's no need for l'Hopital. It's easy to show that (n^2+1)/(2n^2+1) goes to 1/2 as n goes to infinity - just divide the numerator and denomonator by n^2. Since ln is continuous, the sequence converges to ln(1/2) =/= 0 so the series diverges.

dlevanchuk
dlevanchuk is offline
#4
Oct11-09, 04:29 PM
P: 29

convergent/divergent with natural log


thank you very much


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