Question

Hey everyone... can't seem to understand the comparison test.... These convergent and divergent series are really hard for me :/. I always thought that if the bigger series is convergent, then so is the baby one. If the baby one is convergent, then so is the bigger one.

Answer 1

The questions are attached...

Answer 2

@antomic The comparison test states that suppose\[\sum_{n =1}^{\infty}a _{n}\]and\[\sum_{n =1}^{\infty}b _{n}\]are two series with positive terms. If

i). the second series is convergent and the terms of the first sequence are less than or equal to the terms of the second for all n, then the first series is also convergent.
ii). the second series is divergent and the terms of the first sequence are greater than or equal to the terms of the second for all n, then the first series is also divergent.

Example:
Test the series\[\sum_{n =1}^{\infty}\frac{ \ln n }{ n }\]for convergence or divergence.

Solution:
Observe that ln n > 1 for n ≥ 3. Thus\[\frac{ \ln n }{ n }>\frac{ 1 }{ n }\]whenever n ≥ 3

We know that\[\sum_{n =1}^{\infty}\frac{ 1 }{ n }\]is divergent because the power of n is 1 (p-series test). Thus the given series is divergent by the comparison test.

Answer 3

Thank you very much. That was really helpful. I was just wondering, what are the rules for picking the series to compare the original to? Or do you just have to "feel" it?

Answer 4

Sorry my server keeps resetting for some reason which is why it seems like it's taking long. I was was almost finished this time when it restarted. Give me a few minutes to type it up again.

Answer 5

No problem. Don't worry :). i greatly appreciate the help :)

Answer 6

To do the comparison test (not the limit comparison test), compare the given series to the dominant terms of the numerator and the denominator and consider the p-series test.

Example:
Does the series\[\sum_{n =1}^{\infty}\frac{ n ^{2}+1 }{ n ^{4}+1 }\]converge or diverge?

Solution:

First we notice that the dominant term in the numerator is the second power of n, and the dominant term in the denominator is the fourth power of n. Thus we compare the given series to the quotient of the dominant terms. Thus we see that\[\frac{ n ^{2}+1 }{ n ^{4}+1 }<\frac{ n ^{2} }{ n ^{4} }=\frac{ 1 }{ n ^{2} }\]for all n > 1

From this we observe that\[\sum_{n =1}^{\infty}\frac{ 1 }{ n ^{2} }\]is convergent because the power of n > 1 (p-series test).

∴ the given series is convergent by the comparison test.

Answer 7

@antomic do you understand?

Answer 8

Sorry... for some reason it is very laggy... but yes... thank you so much. I think I've got it down now. You are awesome ^_^.

Answer 9

@antomic thank you! Awesome teachers have awesome students so that makes you an awesome student!

Answer 10

haha... man I wish... I'm usually very slow with learning calculus... This is one of the larger hurdles for me... the biggest one was getting through trigonometric substitutions.... I am a slow learner :/... I have to have someone repeatedly nail the material into my brain xD

Answer 11

@antomic you did just fine! Everyone learns at different rates. How fast or slow is irrelevant, what is important is that you persevere!

Answer 12

Thanks again :). I will definitely seek your help if I get stuck again (which should be relatively soon haha).

Answer 13

Have more confidence in your self. let me know whenever you need my help.