# Series cos(x)^n/3^n boundary question

1. Oct 6, 2009

### NastyAccident

1. The problem statement, all variables and given/known data
$$\sum^{infinity}_{n=0}\frac{cos(x)^n}{3^n}$$

Find the values of x for which the series converges.
(z,p)

Find the sum of the series for those values of x.

2. Relevant equations
Geometric Series Sum (a/(1-r))
Geometric Series a*r^n

3. The attempt at a solution
I believe I have b correct, so I'm going to lead with b solution:
$$\sum^{infinity}_{n=0}\frac{cos(x)^{n}}{3^{n}}$$

$$\sum^{infinity}_{n=0}(\frac{cos(x)}{3})^{n}$$
This series is a geometric series a = 1 and r = $$\frac{cos(x)}{3}$$.
Since, $$\left|r\right|=\left|\frac{cos(x)}{3}\right|<1$$, it converges and gives
$$\frac{1}{1-\frac{cos(x)}{3}}$$

Attempt at a.
Since the series starts at n = 0, the first term, is always going to be 1. Find the values of x for which the series converges. Thus,
$$\left|\frac{cos(x)}{3}\right|<1$$

$$-1<\frac{cos(x)}{3}<1$$

$$-3<cos(x)<3$$

$$arccos(-3)<x<arccos(3)$$

But, the issue is the the domain of arccos(x) is -1<x<1... I'm not necessarily sure what I'm doing wrong for this particular problem.

I've tried -1/3 (since the range of cosine is -1 to 1) and -3, both were incorrect.

As always any help or further explanation is welcome and will be greatly appreciated!

Sincerely,

NastyAccident

2. Oct 6, 2009

### lanedance

hi NastyAccident, may computer is being a bit silly with tex to read, but hope this helps

first does the series below converge?
$$\sum^{infinity}_{n=0}\frac{1}{3^{n}}$$

then maybe have a think about absolute convergence & the absolute value of cos

3. Oct 6, 2009

### HallsofIvy

Staff Emeritus
What you did at first is correct and completely answers the question: that is a geometric series in cos(x) and converges as long as -1< cos(x)/3< 1 which is the same as -3< cos(x)< 3. Since the range of cosine is only -1 to 1, that converges for all x and the sum is
$$\frac{1}{1- \frac{cos(x)}{3}}= \frac{3}{3- cos(x)}$$

4. Oct 6, 2009

### NastyAccident

Ahh! So, I should of stopped on the step before I took the arccos of each side.

Now, a new question that I have is why did you just leave it as cos(x)? The normal thought is to create an inequality like -1<x<1 that has x solely in the middle. Since there was a trigonometry function, is it just custom to have an inequality like -1<cos(x)<1?

From after that step, I can easily understand everything else: such as, cos(x) range is from -1 to 1 and range for this particular series would be [-1/3,1/3]. This fits inside the -3<cos(x)<3 inequality.

Many thanks for your help Ivy!

Sincerely,

NastyAccident

Last edited: Oct 6, 2009