# Using the limit comparison test to prove conv or div

• Randall
In summary, the limit comparison test says that if you have two positive series, sum An and sum Bn, let C=lim n to infinity of An/Bn and (i) if 0< C < infinity; then both series converge or both diverge; (ii) if C = 0 and sum Bn converges, so does sum An; (iii) if C = infinity and sum Bn diverges, so does sum An.

## Homework Statement

Use the limit comparison test to prove convergence or divergence for the series sum from n=1 to infinity for ((5n^3)+1)/((2^n)((n^3)+n+1))

## Homework Equations

The limit comparison test says that if you have two positive series, sum An and sum Bn, let C=lim n to infinity of An/Bn and (i) if 0< C < infinity; then both series converge or both diverge; (ii) if C = 0 and sum Bn converges, so does sum An; (iii) if C = infinity and sum Bn diverges, so does sum An.

## The Attempt at a Solution

For An, I used L'Hospital's rule (twice) to get lim as n to infinity of 30/infinity, which equals 0. For Bn, I chose Bn = 2^n, which is a divergent geometric series with an r of 2 (>1). This didn't help me apply the limit comparison test, so I changed my Bn to the convergent geometric series 0.5^n, which has an r of 0.5 (<1). This also didn't help me apply the limit comparison test. Can someone help me please? Thanks so much in advance.

You don't take the the limits of ##A_n## and ##B_n## separately for the limit comparison test. What you want to try for##B_n## is not ##2^n## but ##\frac 1 {2^n}## which you have in your series. So you should be calculating the limit of $$\frac{\frac{5n^3+1}{(2^n)(n^3+n+1)}}{\frac 1 {2^n} }$$Simplify that and work out that limit. You shouldn't need L'Hospital's rule to work that limit.

Randall said:
so I changed my Bn to the convergent geometric series 0.5^n, which has an r of 0.5 (<1). This also didn't help me apply the limit comparison test. .

Show why didn't it work to use $B_n = 0.5^n = \frac{1}{2^n}$

Hi Stephen: so when I used Bn = 0.5^n, and then analyzed it as n goes to infinity, I got Bn = 0. And, since I already determined that An=0, that makes my C = 0/0, which is undefined, yes? So I can't apply the limit comparison test for Bn being a convergent series.

Randall said:
Hi Stephen: so when I used Bn = 0.5^n, and then analyzed it as n goes to infinity, I got Bn = 0. And, since I already determined that An=0, that makes my C = 0/0, which is undefined, yes? So I can't apply the limit comparison test for Bn being a convergent series.
Did you read the first sentence of my post #2? Or post #2 at all?

Randall said:
Hi Stephen: so when I used Bn = 0.5^n, and then analyzed it as n goes to infinity, I got Bn = 0. And, since I already determined that An=0, that makes my C = 0/0, which is undefined, yes? .

No. If that were true the l'Hosptials rule wouldn't work. Reduce the fraction $\frac{A_n}{B_n}$ and take the limit of $\frac{A_n}{B_n}$ using l'Hospital's rule.

LCKurtz: Ok thanks I tried that but still had to use L'Hospital's rule to evaluate the limit as n goes to infinity. For Bn, I ended up with lim n to inf of 15 / 6n = 15/infinity = 0. So, since An = 0, I get a C of 0/0, which is undefined yes, or is C = 0? Thanks.

LCKurtz, no I had not read it when I was responding to Stephen, but I have read it since and have replied to you.

You need to show your work. Show us what you took the limit of and how you did it.

Randall said:
LCKurtz: Ok thanks I tried that but still had to use L'Hospital's rule to evaluate the limit as n goes to infinity. For Bn, I ended up with lim n to inf of 15 / 6n = 15/infinity = 0. So, since An = 0, I get a C of 0/0, which is undefined yes, or is C = 0? Thanks.
What LCKurtz meant in the first sentence of post 2 is that
$$C = \lim_{n \to \infty} \frac{A_n}{B_n} \ne \frac {\lim_{n \to \infty} A_n}{\lim_{n \to \infty} B_n}.$$ From what you've written, it sounds like you're finding the quotient of the two limits, which isn't what you should be doing. You should be calculating the limit of the quotient ##A_n/B_n##.

To your quoted formulation of the comparison test I prefer the one in the blue box here http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx or the one in the first lines here http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx .

Because you are not asked to find a limit but only establish that there is one, the comparison test allows you to work with < instead of = , which allows you considerable freedom in eliminating awkward terms you don't want and replacing your ((5n^3)+1)/((2^n)((n^3)+n+1))
with an easier and more obviously convergent expression.