Limit Comparison Test for Divergence: Solving Problems with a_n/b_n

[MillerGenuine]

Homework Statement

<br /> \sum_{n=1}^\infty \frac{1+4^n}{1+3^n}<br /> where a_n= \frac{1+4^n}{1+3^n} and b_n= \frac{4^n}{3^n}<br />

Homework Equations

I know how to do this problem, you take the limit as "n" goes to infinity of a_n/b_n ... which after a good amount of algebra ends up being 1. which is greater than zero. once i do this I go on to the next step which is...

<br /> \sum_{n=1}^\infty \frac{4^n}{3^n}<br />

ok so now once i get this I see that it is a geometric series where r= 4/3 which is greater than 1..which means the series Diverges by limit comparison test.


Easy enough probelm..but my question is what's with the hole limit comparison test portion? can't i just see my b_n= 4^n/3^n and therefore seeing its a geometric series, skipping the limit test all together and moving straight to Direct Comparison Test.. the second step i showed above. I tried doing this (skipping limit test) for a few problems in my book and they ended up being the same...so please tell me..whats with the foreplay of finding the limit, when it seems i can just go straight to Direct Comparison test by using my b_n to see if its a Geometric series or P-series?

 

[Mark44]

Homework Statement

<br /> \sum_{n=1}^\infty \frac{1+4^n}{1+3^n}<br /> where a_n= \frac{1+4^n}{1+3^n} and b_n= \frac{4^n}{3^n}<br />

Homework Equations

I know how to do this problem, you take the limit as "n" goes to infinity of a_n/b_n ... which after a good amount of algebra ends up being 1. which is greater than zero. once i do this I go on to the next step which is...

<br /> \sum_{n=1}^\infty \frac{4^n}{3^n}<br />

ok so now once i get this I see that it is a geometric series where r= 4/3 which is greater than 1..which means the series Diverges by limit comparison test.


Easy enough probelm..but my question is what's with the hole limit comparison test portion? can't i just see my b_n= 4^n/3^n and therefore seeing its a geometric series, skipping the limit test all together and moving straight to Direct Comparison Test.. the second step i showed above. I tried doing this (skipping limit test) for a few problems in my book and they ended up being the same...so please tell me..whats with the foreplay of finding the limit, when it seems i can just go straight to Direct Comparison test by using my b_n to see if its a Geometric series or P-series?

I think that you are misunderstanding the direct comparison test. In the comparison test, if a_n is the general term of the series you're testing, and b_n is the general term of the series you're testing against, if you think your series diverges, then you need to show that a_n > b_n, where b_n is the general term of a divergent series.

OTOH, if you think your series converges, then you need to show that a_n < b_n, where this time b_n is the general term of a convergent series.

For the series in this problem, can you show that
\frac{1 + 4^n}{1 + 3^n} &gt; \frac{4^n}{3^n}?

Maybe you're thinking you need to use both tests - you don't. Many times it's more convenient to use the limit comparison test, since it can be difficult to establish the inequality of the direct comparison test.

 

[MillerGenuine]

For the series in this problem, can you show that \frac{1 + 4^n}{1 + 3^n} &gt; \frac{4^n}{3^n}

?

What do you mean by "can i show" ?

Maybe you're thinking you need to use both tests - you don't.

For all the limit comparison test problems in my book, as well as problems shown in lecture, It always shows the problem worked by first showing (proving) limit a_n/b_n > 0
Then once this is shown they always seem to take the summation of b_n and show that it is either a p-series/g-series/harmonic..as if this is the ultimate conclusion as to why the series converges (as i showed in the above example). Making it seem like taking the limit is unnecessary.

 

[Mark44]

What I mean by "can you show ..." is that when you use direct comparison, you need to show that your series is term-by-term larger than some known divergent series, or term-by-term smaller than some known convergent series. The other two possibilities (your series is smaller than a known divergent series, or your series is larger than a known convergent series) don't provide any guidance.

Both kinds of tests -- limit comparison and direct comparison -- require that you compare the series you're investigating with a series whose behavior is known. When your text shows that lim a_n/b_n > 0, it's only half done. All this shows is that the series being investigated has the same behavior as \sum b_n. To complete the work, they are showing that the series being compared to converges or diverges, thus the series they're actually working with does the same.

 

[MillerGenuine]

OHH! Ok that makes sense now. So your showing that your series has the same behavior as b_n, and then you must show that your b_n is convergent in order to make the comparison that a_n is convergent as well. I guess its in the title, Direct comparison test. wow ok perfect. Thank you