# McLauren Series

1. Mar 31, 2009

### nns91

1. The problem statement, all variables and given/known data

The MacLaurin series for ln(1/1-x) is sum of x^n/n with interval of convergence -1<= x <1

a. Find the Maclaurin series for ln(1/1+3x) and determine the interval of convergence
b. Find the value of $$\sum$$ (-1)^n/n from n=1 to infinity.
c. Give a value of p such that $$\sum$$ (-1)^n/ n^p converges but $$\sum$$ of 1/n^(2p) diverges. Give reasons why your value of p is correct.
d. Give a value of p such that $$\sum$$1/n^p diverges but $$\sum$$1/n^(2p) converges. Give reasons why your value of p is correct.

2. Relevant equations

Limit test

3. The attempt at a solution

a. So I guess the answer for the series is $$\sum$$(-1)3x^n/n. Am I right ?
b. How do I do this one ?
c. I am not sure how to do this one.
d. I think 1/n^p diverges when p<1. For p<1, 2p<2 so 1/n^2p < 1/n^2 which is a convergent series. So my p is correct. Am I right ?

2. Mar 31, 2009

### nns91

I think I did part d wrong to. Right ?

3. Apr 1, 2009

### HallsofIvy

Staff Emeritus
No, you are not. HOW did you "guess the answer"?

Try something!

You are not doing the right problem. You were asked to find p such that $\sum (-1)^n/n^p$ converges and $\sum 1/n^p$ diverges.

4. Apr 1, 2009

### nns91

a. I actually did not guess, I multiplied x by 3 and negated it. If am wrong, how should I do it then ?

d. Oh, I see. I misunderstood the problem. Then p should be 1 then, right ? since 1/n diverges but 1/n^2 converges

5. Apr 1, 2009

### lanedance

just to conceptualise it, try re-writing, first from your problem description you have (i haven't checked this...):
$$ln(\frac{1}{1-x}) = \sum \frac{x^n}{n}$$

and you want to find
$$ln(\frac{1}{1+3y}) =$$

can you find x = f(y) and substitute this into you original taylor expansion?

d) sounds reasonable if p is an integer, can you expain why though? think convergence tests

6. Apr 1, 2009

### nns91

. How should I do part b ?

7. Apr 1, 2009

### rootX

ln(1/1+3x) = ln(1/1-(-3x))

You had ln(1/1-x) and now you are multiplying -3 to x