What values of x make this series converge?

[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 don't 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 can't 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? I am 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

 

[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?

 

[Galileo]

You correctly recognized it as a geometric series.

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

 

[Alem2000]

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

 

[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|&lt;1 and r=\frac{x}{3} for what values of x does the series converge?

 

[Alem2000]

for \frac{x}{3}&lt;1 i don't know how to put in the absoulute value signs so imagine that fraction has an abs sign and that comes out to be x&lt;3 Yippy! is that my answer? Seems to be correct

 

[Galileo]

Yep
Since |r|=|\frac{x}{3}|
|r|&lt;1 \iff |x/3|&lt;1 \iff |x|&lt;3

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

 

[HallsofIvy]

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}&lt; 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}&lt; 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.

 

[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