# Understanding series?

1. Nov 8, 2004

### Alem2000

Understanding series?????

The question in the book states Find the values of x for which the series converges, Find the sum of the series for those valuese of x

this is the series $$\sum_{n=1}^{\infty}\frac{x^n}{3^n}$$

first of all I dont even really understand what its saying, is it saying find the sum and set x equal to it.? The first thing I did was write out the first few terms $$[\frac{x}{3}],[\frac{x^2}{9}],[\frac{x^3}{27}]$$ so this thing is geometric and $$a=\frac{x}{3}$$ with the ratio $$r=\frac{x}{3}$$ am I assuming that that the ratio is less than one b/c if not you cant even go anyfurther can you? well anyway after that by using the $$\frac{a}{1-r}$$ theorm I get $$\frac{x}{3-x}$$ where do I go from here? Im pretty confused If this was a series with real numerical values I would have been done with the question b/c I have already fournd the sum...but what next...should there be another function that I set this equal to to get the "values of x" that the book wants

Last edited: Nov 8, 2004
2. Nov 8, 2004

### philosophking

I unfortunately haven't done series for a couple years, but I definitely know that r<1 in order for the series to converge, which would lead me to believe that 0<x<3 maybe? or 0 < or = x < 3 ?

I guess from there you would have x^n/3^n = (x/3)^n, where you could then apply your summation of geometric series rule, where r = x/3 (given any x in the interval defined above), and a_0 = x/3. So,

(x/3)/ [1-(x/3)] = x/(3-x), for x >/= 0, x<3. Maybe?

3. Nov 8, 2004

### Galileo

You correctly recognized it as a geometric series.

For which values of r does $\sum_{n=0}^{\infty}r^n$ converge?

4. Nov 8, 2004

### Alem2000

those whose absolute values are less then one...should i set my equation equal to 1?

5. Nov 8, 2004

### Galileo

You know a geometric series converges for |r|<1.
You also identified your series as a geometric series with $r=\frac{x}{3}$.

If the series converges for $|r|<1$ and $r=\frac{x}{3}$ for what values of x does the series converge?

6. Nov 8, 2004

### Alem2000

for $$\frac{x}{3}<1$$ i dont know how to put in the absoulute value signs so imagine that fraction has an abs sign and that comes out to be $$x<3$$ :rofl: Yippy! is that my answer? Seems to be correct

7. Nov 9, 2004

### Galileo

Yep
Since $|r|=|\frac{x}{3}|$
$$|r|<1 \iff |x/3|<1 \iff |x|<3$$

I just use pipelines for absolute values. (shift+backslash)

8. Nov 9, 2004

### HallsofIvy

Staff Emeritus
It's relatively easy to show that any power series (i.e. anything of the form $\Sigma a_n x^n$) has a "radius of convergence": a number r such that the series converges absolutely for |x| < r, diverges for |x|> r and may or may not converge for |x|= r. Often the simplest way to find the radius of convergence is to use the "ratio test"- a series $\Sigma a_n$ converges if $lim\frac{a_{n+1}}{a_n}< 1$ and diverges if that limit is larger than 1.

In this particular example, $\sum_{n=1}^{\infty}\frac{x^n}{3^n}$, the ratio becomes $\frac{|x^{n+1}|}{3^{n+1}}\frac{3^n}{|x^n|}= \frac{|x|}{3}$. The series converges if $\frac{|x|}{3}< 1$ or |x|< 3. Of course, as was pointed out before, this is really a geometric series.

Another example might be $\Sigma\frac{x^n}{n}$. Now the ratio becomes $\frac{|x^{n+1}|}{n+1}\frac{n}{|x^n|}= \frac{n||x}{n+1}$. The limit of that as n goes to infinity is just |x| so we must have |x|< 1. The radius of convergence is 1.

Last edited: Nov 9, 2004
9. Nov 9, 2004

### Alem2000

$$\frac{|x^{n+1}|}{n+1}\frac{n}{|x^n|}= \frac{n|x|}{n+1}$$

is that what you ment? I think your latex was wrong