1. power series

write a power series for f(x) = 3/(2-4x). write the series in closed form. give the interval of convergence.

3/(2-4x)

= (3/2)[1/(1-2x)]

sum n=o to oo (3/2)(2^n)(x^n)

how do i get the interval of convergence?

2. $\displaystyle \frac {3}{2} \sum_{n=0}^{ \infty} (2x)^n$

We need not consider the constant $\displaystyle \frac {3}{2}$

By the Root Test, we have:

$\displaystyle \lim_{n \to \infty} \left| (2x)^n \right|^{ \frac {1}{n}} = \lim_{n \to \infty} \left|2x \right|$

The series converges for $\displaystyle \left| 2x \right| = 2 \left| x \right| < 1 \Rightarrow |x| < \frac {1}{2}$

So the radius of convergence is $\displaystyle \frac {1}{2}$

Test the endpoints:

If $\displaystyle x = \frac {1}{2}$

$\displaystyle \Rightarrow \sum (2x)^n = \sum 1^n$ which diverges by the test for divergence

If $\displaystyle x = - \frac {1}{2}$

$\displaystyle \Rightarrow \sum (2x)^n = \sum (-1)^n$ which diverges by the test for divergence

so the Interval of convergence is $\displaystyle \left( - \frac {1}{2} , \frac {1}{2} \right)$

3. how did the exponent 1/n get there?

oh never mind its root test not ratio...sorry. we werent taught to use that test so i it didnt come to my mind at first...

4. Originally Posted by jeph
how did the exponent 1/n get there?
i used the root test, you always put an exponent 1/n.

you would get the same result using the ratio test, but since the exponent of 2x was just n, i decided to use the root test to get rid of the power, instead of going through the root test and having to form a fraction and all that stuff

the ratio test will give the same result, now that you know the procedure to find the IOC, try it using the ratio test

5. Here is a weaker version of the root canal test. The strong version is based on the notion of a "limsup" which is not convered in any Calculus course.

Given a series $\displaystyle \sum_{n=1}^{\infty}a_n$. And given the limit $\displaystyle L = \lim |a_n|^{1/n}$ (if it exists or infinite). We have the following cases:

If $\displaystyle L<1$ then the series converges absolutely.

If $\displaystyle L>1 \mbox{ or }L=+\infty$ then the series diverges.

If $\displaystyle L=1$ the test doth not work.

This is a beautiful test, which should be taken advantage of as much as possible. I like it more than the ratio test. (In fact it is stronger!)

6. Originally Posted by ThePerfectHacker
Here is a weaker version of the root canal test. The strong version is based on the notion of a "limsup" which is not convered in any Calculus course.

This is a beautiful test, which should be taken advantage of as much as possible. I like it more than the ratio test. (In fact it is stronger!)
lol, why call it the "root canal" test then?

i always have trouble applying it when factorials are involved, i think i did it once though.

7. Originally Posted by Jhevon

lol, why call it the "root canal" test then?

i always have trouble applying it when factorials are involved, i think i did it once though.
I found the following useful. (Chapter 12)

1) $\displaystyle \lim (n!)^{1/n} = +\infty$
2) $\displaystyle \lim \frac{1}{n} (n!)^{1/n} = \frac{1}{e}$

I can derive it. But I remember I derived it for you before.

8. Originally Posted by ThePerfectHacker
I found the following useful. (Chapter 12)

1) $\displaystyle \lim (n!)^{1/n} = +\infty$
2) $\displaystyle \lim \frac{1}{n} (n!)^{1/n} = \frac{1}{e}$

I can derive it. But I remember I derived it for you before.
Yes, i remember. i believe we derived it by the Ratio test though