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Series and Sequences

  • #1
Hope all of you had a good July 4th (for those that celebrate it)!

Anyways, I'm trying to figure out series and sequences. I'm using Stewarts' Single Variable Calculus: Early Transcendentals, 6th. ed.

For instance, under Section 11.4, the Comparison Tests, I don't understand how one arrives at bn.

Example:

[tex]\Sigma[/tex][tex]^{\infty}_{n=1}[/tex][tex]\frac{1}{2^{n}-1}[/tex]

The book then proceeds to state:

an = [tex]\frac{1}{2^{n}-1}[/tex], which I understand because it's written as [tex]\Sigma[/tex]an.

However, the book then proceeds to state that:

bn = [tex]\frac{1}{2^{n}}[/tex], which I have no idea how they got there. Do I just remove all constants from an? Or do I remove all variables that are not attached by an "n"?


Also, my Calculus 1 is a bit rusty, but how exactly does (excerpted from p.717, Example 4 of Stewarts)

[tex]\frac{(n+1)^{3}}{3^{n+1}}[/tex] [tex]\times[/tex] [tex]\frac{3^{n}}{n^{3}}[/tex]



=[tex]\frac{1}{3}[/tex]([tex]\frac{n+1}{n}[/tex])3

=[tex]\frac{1}{3}[/tex](1+[tex]\frac{1}{n}[/tex])3

=[tex]\frac{1}{3}[/tex]
 

Answers and Replies

  • #2
29
0
i don't have a copy of that book, so here's a conjecture (going by the looks of it):
that's the ratio test. you determine the ratio of (n+1)th term to the n th term as n approaches infinity, and conclude depending on whether the value obtained is greater or less that 1, the series is cgt or dvt. in this case it turns out to be 1/3.
 
  • #3
761
13
Hope all of you had a good July 4th (for those that celebrate it)!

Anyways, I'm trying to figure out series and sequences. I'm using Stewarts' Single Variable Calculus: Early Transcendentals, 6th. ed.

For instance, under Section 11.4, the Comparison Tests, I don't understand how one arrives at bn.
It's just implementing the comparison test:

11.PNG
 
  • #4
  • #5
I just skimmed your first post... I think this is relevant though:
B sub n is just some other series that was (almost) arbitrarily picked because it converges.
 
  • #6
You can use the limit comparison test. I don't know a good mathematical way to explain this, but when I learned that test, my instructor called bn the "simpler version" of an. As it was stated earlier, we know that [tex]\sum^{\infty}_{n=0}[/tex][tex]\frac{1}{2^{n}}[/tex] converges. So, the limit comparison test says that if lim[tex]_{n\rightarrow\infty}[/tex][tex]\frac{a_{n}}{b_{n}}[/tex]=c, and c>0, then both an and bn converge or both diverge, if bn converges, so does an and the same goes for divergence.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,777
911
[tex]\frac{1}{2^n-1}> \frac{1}{2^n}[/tex]
because the numerator on the left is smaller than the numerator on the right. However, that really doesn't help because you want to show this converges so you need "<" not ">" to a convergent series.

You might say [tex]\frac{1}{2^n-1}< \frac{2}{2^n}[/tex]
and then argue that since [tex]2\sum_{n=0}^\infty \frac{1}{2^n}[/tex] converges, so does [tex]\sum_{n=0}^\infty \frac{1}{2^n-1}[/tex]

and, of course,
[tex]\frac{(n+1)^3}{3^{n+1}}\frac{3^n}{n^3}= \frac{3^n}{3^{n+1}}\left(\frac{n+1}{n}\right)^3[/tex]
[tex]= \frac{1}{3}\left(1+ \frac{1}{n}\right)^3[/tex]
 

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