Question

Help with determining convergence test. I cant figure out which test to use. I tried the Ratio test and it turns out realllllllllly ugly. I have a feeling that I am going to have to use comparison, but I am kinda stuck.. Here is the series..

\[\sum_{n=1}^{\infty}(2n+6)/\sqrt{6n^4+4n+4}\]

Answer 1

i would say sqrt(6n^4 is almost like sqrt3n^2 and one n would cancel and it would like like 1/n which diverges

Answer 2

but i could be very very wrong

Answer 3

Divide the top and bottom by n, so the top becomes 2+6/n and on the bottom (by dividing by n=sqrt(n^2)) you get sqrt(6n^2+4/n+4/n^2). As n goes to infinity the bottom has a positive n and the top has one factor less, meaning after each iteration (increasing n by 1) the sequence is getting smaller. It's converging, I believe.

Answer 4

lim would = 0 which would be convergent according to the test for divergence

Answer 5

it diverges...NoxiPro thinking is spot on. One could do a comparison test to prove this.

Answer 6

'Sum::div: Sum does not converge'

Mathematica 8.0 agrees with me ;)

Answer 7

he was on til after the 1/n part. As you go to infinity with something that has 1/n the sequence gets smaller with each iteration, which means it will converge to a point..

Answer 8

thanks zarkon.

Answer 9

1/n diverges because it is a harmonic series

Answer 10

he was doing a quick comparison with 1/n

\[\sum_{n=1}^\infty\frac{1}{n}\] diverges

Answer 11

Yes: http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Answer 12

But what would i use to compare it to do comparison test

Answer 13

why does that diverge? That would be 1+.5+.25+.125...right..converging to 1..?

Answer 14

For ease sake im going to use A for num and B for denom.. Would I just say a/b>1/b and then find the limit as n->infin?

Answer 15

well, to 2, not one..

Answer 16

physopholy: Read the wikipedia article, there is a very easy to understand proof of divergence of the harmonic series.

Answer 17

for \[n\geq1\]

\[\frac{2n+6}{\sqrt{6n^4+4n+4}}\geq\frac{2n}{\sqrt{6n^4+4n+4}}\geq\frac{2n}{\sqrt{6n^4+4n^4+4n^4}}=\frac{2n}{\sqrt{14n^4}}\]

\[=\frac{2n}{\sqrt{14}n^2}=\frac{2}{\sqrt{14}}\frac{1}{n}\]

Answer 18

1 +(1/2+1/3+1/4+1/5)=1+(60+40+30+24/120)>1
so its like 1+1 and you keep grouping terms to get >1 and since you go forever you can always collect terms to add "1"

Answer 19

Do a comparison with an obviously diverging series (from wiki):
\[\begin{align}
& 1 \;\;+\;\; \frac{1}{2} \;\;+\;\; \frac{1}{3} \,+\, \frac{1}{4} \;\;+\;\; \frac{1}{5} \,+\, \frac{1}{6} \,+\, \frac{1}{7} \,+\, \frac{1}{8} \;\;+\;\; \frac{1}{9} \,+\, \cdots \\>\;\;\; & 1 \;\;+\;\; \frac{1}{2} \;\;+\;\; \frac{1}{4} \,+\, \frac{1}{4} \;\;+\;\; \frac{1}{8} \,+\, \frac{1}{8} \,+\, \frac{1}{8} \,+\, \frac{1}{8} \;\;+\;\; \frac{1}{16} \,+\, \cdots
\end{align}\]

Answer 20

^thats the one i saw in class

Answer 21

Well thats a tricky one.. I guess I was on the right track with the comparison but it was a lot of jumping through hoops for the right series to compare it to lol, thanks all

Answer 22

ok to get a quick refernce. 1/n diverges but 1/n^2=pi^2/6 which converges use those to do your comparison test

Answer 23

o.o how does 1/n^2 = pi^2/6? or is that what you get when you converge n^2

Answer 24

thisone if im not mistaken uses something with euler. but the proof was more complicated than 1/n

Answer 25

I know that most likely, when you compare, you want to a p-series to use.. where if p>1 it converges and when \[p \le1\] it diverges.. But sometimes I have trouble determining the easiest p-series

Answer 26

p series being 1/n^p

Answer 27

it takes doing lots and lots of problems to recognize what to use

Answer 28

the down falls of a summer semester :P

Answer 29

yeah i would advise not doing calc 2 in summer

Answer 30

LOL well it is too late now, only 2 weeks left

Answer 31

your fine if youve made it to seq your good