Question

Using series? Help?

Answer 1

the first one we can compare it sum 1/n
since 1/n > 1/root (n^2+1)

Answer 2

Okay sounds good so far :)

Answer 3

1/n is harmonic series that diverges
therefore the series must diverge by the comparison test

Answer 4

Yes, I understand that :)

Answer 5

What about the second one?

Answer 6

hmm the second
how about you do what they asked limit comparison test
\[\lim_{n\to \infty }\frac{a_{n+1}}{a_n}\]

Answer 7

so what will that be?

Answer 8

well the first i actually did the comparison test rather than limit comparison test

Answer 9

what a_n+1?

Answer 10

n+3?

Answer 11

hmm n+3 what do yo mean?

Answer 12

well the top will be n+3
but i'm asking the whole thing

Answer 13

I'm not sure... heh. .he.h..he.h.

Answer 14

Would you just add one to the original equation?

Answer 15

well just replace n
with n+1
\[a_n=\frac{n+2}{(n+1)^2}\]

Answer 16

(n+3) / (n+2) ^2

Answer 17

I get you bro

Answer 18

yes good!

Answer 19

no divide that be a_n
because we want a_n+1/a_n

Answer 20

How does that help? :)

Answer 21

we will take the limit!

Answer 22

if the limit is some number L then the series must converge
if it 0 or infinity no conclusion

Answer 23

that is the limit comparison test tells us

Answer 24

I got that the limit converge to 1?

Answer 25

then the series must converge

Answer 26

Does the property state ( the limit comparison) that the series may both converge OR diverge?

Answer 27

But what do we compare it to?

Answer 28

your question is somehow off
they are asking to use those series to check the convergence of the wanted series
but they want it with limit comparison test
that should of been comparison test not limit

Answer 29

Or I think they are asking for the limit we compare it to... then we can use the limit comparison theorem? All the series they provide can be determined if they are convergent or divergent by the p-series test.

Answer 30

Wait, I'm confused xD

Answer 31

I'm not getting your point here
but the limit comparison test tells us that if the limit is finite and is L
a_n approax equal La_n+1
if one converges so does the other
by the boundness theorem i believe

Answer 32

here seems to me they want to do comparison test

Answer 33

reference
http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx

Answer 34

Okay your way works too! Yeah

Answer 35

so you want to do regular comparison test? or limit comparison test
in fact limit comparison test is much powerful than the regular
all you need is to find the limit if it finite and positive then the series converge

Answer 36

at any rate first conveges
let's the second with what series can we compare it to

Answer 37

\[\sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}}\] does not converge

Answer 38

oh you know i see
it is my mistake actually , the way i did this was different
so the first we would have done
\[\lim \frac{\frac{1}{\sqrt{n^2+1}}}{1/n}\]

Answer 39

of course does not converge harmonic series
did i say the opposite? lol

Answer 40

look right above my post

Answer 41

hehe slept over that, i did say from the start that it diverges though
thanks for the correction

Answer 42

the second as will i would compare it with 1/n

Answer 43

so divergent as a result

Answer 44

the third
with 1/n^3/2
what I'm really doing here is looking at the leading terms of top and bottom and decide which series would work

Answer 45

but doesn't work all the time i believe
here 1/n^3/2 is convergent series (3/2>1) p series if you recall

Answer 46

the real work you should take the limit as i did above before you draw the conclusion though

Answer 47

just giving you where this is going...

Answer 48

Yes, I did... Thank you very much, I think I got it now (: