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Convergent/divergent with natural logby dlevanchuk
Tags: natural 
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#1
Oct1109, 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?? 


#2
Oct1109, 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? 


#3
Oct1109, 04:08 PM

PF Gold
P: 867

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.



#4
Oct1109, 04:29 PM

P: 29

Convergent/divergent with natural log
thank you very much



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