# Series and Sequences

1. Jul 5, 2008

### sharkshockey

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:

$$\Sigma$$$$^{\infty}_{n=1}$$$$\frac{1}{2^{n}-1}$$

The book then proceeds to state:

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

However, the book then proceeds to state that:

bn = $$\frac{1}{2^{n}}$$, 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)

$$\frac{(n+1)^{3}}{3^{n+1}}$$ $$\times$$ $$\frac{3^{n}}{n^{3}}$$

=$$\frac{1}{3}$$($$\frac{n+1}{n}$$)3

=$$\frac{1}{3}$$(1+$$\frac{1}{n}$$)3

=$$\frac{1}{3}$$

2. Jul 5, 2008

### sihag

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. Jul 6, 2008

### dirk_mec1

It's just implementing the comparison test:

4. Jul 6, 2008

### sharkshockey

Error: Does not compute.

Sorry, I'm still confused

5. Jul 7, 2008

### scorpion990

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. Aug 13, 2008

### BoundByAxioms

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 $$\sum^{\infty}_{n=0}$$$$\frac{1}{2^{n}}$$ converges. So, the limit comparison test says that if lim$$_{n\rightarrow\infty}$$$$\frac{a_{n}}{b_{n}}$$=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. Aug 14, 2008

### HallsofIvy

Staff Emeritus
$$\frac{1}{2^n-1}> \frac{1}{2^n}$$
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 $$\frac{1}{2^n-1}< \frac{2}{2^n}$$
and then argue that since $$2\sum_{n=0}^\infty \frac{1}{2^n}$$ converges, so does $$\sum_{n=0}^\infty \frac{1}{2^n-1}$$

and, of course,
$$\frac{(n+1)^3}{3^{n+1}}\frac{3^n}{n^3}= \frac{3^n}{3^{n+1}}\left(\frac{n+1}{n}\right)^3$$
$$= \frac{1}{3}\left(1+ \frac{1}{n}\right)^3$$