# Limit comparison test assistance needed please

• Randall
This is not an appropriate way to post work here in the Precalculus forums. If you continue to post messy images instead of typing your work, you will be suspended from the forums.
Randall

## Homework Statement

Use the limit comparison test to check for convergence or divergence: Sum from n=1 to infinity of ((2n)^2+5)^-3

## Homework Equations

let lim n to infinity of An/Bn = c
1) if 0<c<infinity then either both converge or both diverge
2) if c=0 and sum Bn converges, so does sum An
3) if c=infinity and sum Bn diverges, so does sum An

## The Attempt at a Solution

see attached - I'm a little confused. It seems that it would be easy to pick a comparison series to "force" the answer that you want...

#### Attachments

• limit_comparison_test.jpg
46 KB · Views: 399
Chosing $b_n = 1 + \frac{1}{\sqrt[3]{n} }$ doesn't let you draw any conclusion because $\sum_{n=1}^\infty b_n$ doesn't converge.

Doesn't it converge to 1? 1 + 0 = 1? If not, why not? And also what should I choose instead to be bn? Thanks

Randall said:
Doesn't it converge to 1? 1 + 0 = 1? If not, why not? And also what should I choose instead to be bn? Thanks

Don't confuse the convergence of the sequence $\lim_{n \rightarrow \infty} b_n = 1$ with the convergence of the series $\sum_{n=1}^\infty b_n$. The part 2 in your statement of the limit comparison tests talks about $b_n$ being summable.

My guess is $b_n = (\frac{1}{n^2})^{-3}$.

Edit: Try $b_n = (n^2)^{-3}$

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That looks like a photo taken with i-pad - for presentation the app mentioned below is almost no extra trouble and makes easier on the eye. DocScan HD

LCKurtz said:
They are the same ##L##. By the way, posting images instead of typing your work is frowned on in these forums. That goes double for images that are posted sideways.

"But if you do do that, there are apps, at least one I know called DocScan HD which if you are taking these with I-pad or similar cleans them up from the hard-to-read grey-yellow."

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If you find it very difficult to prove convergence, hedge your bets and try to test for divergence. I would compare with something like 1/n (which I hope you know is divergent.)

Thread closed. Please stop posting messy images of your work. Most of what you posted has been crossed out, and much of the rest is difficult to read.

## 1. What is the limit comparison test?

The limit comparison test is a method used in mathematics to determine the convergence or divergence of a series. It involves comparing the given series to a known series with known convergence or divergence.

## 2. How do I use the limit comparison test?

To use the limit comparison test, you first need to find a known series with known convergence or divergence that is similar to the given series. Then, you take the limit of the ratio of the terms of the two series. If the limit is a positive number, then the two series have the same convergence or divergence. If the limit is 0 or infinity, then the two series have opposite convergence or divergence.

## 3. When should I use the limit comparison test?

The limit comparison test is typically used when the series involves fractions or exponentials, and the terms in the series are difficult to simplify. It is also useful when the ratio test cannot be used due to the presence of factorial terms.

## 4. Can the limit comparison test be used to prove convergence?

Yes, the limit comparison test can be used to prove convergence. If the limit of the ratio of the terms is a positive number, then the two series have the same convergence. This means that if the known series converges, then the given series also converges.

## 5. What is the difference between the limit comparison test and the ratio test?

The main difference between the limit comparison test and the ratio test is the way the limit is taken. In the ratio test, the limit is taken of the absolute value of the ratio of the terms, while in the limit comparison test, the limit is taken of the ratio of the terms itself. Additionally, the ratio test can only determine absolute convergence, while the limit comparison test can determine both absolute and conditional convergence.

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