find the radius of convergence of the power series......

Question

chrisplusian · Answer

I was given a function and when I used the ratio test to find the radius of convergence it came out generically that $$\lim_{n \rightarrow \infty}\left| \frac{ a _{n+1} }{ a _{n} } \right|=0$$ doesn't this mean that there exists just one point where there is convergence for the power series? In my textbook they say there are three possible cases for the radius of convergence:
1) There is one distinct point where the power series converges.
2) There is an interval "r" that can include the endpoints or not that the power series converges on.
3)the power series converges for all real numbers. 
Then they do three distinct examples. They show where when you use the ratio test the result is infinity, in that case they said the power series converges for all real numbers. Then they did one where the ratio test gives the result as zero, and say that it converges only at the point where the series is centered, and last they did an example where they showed the radius was an interval. Now I am doing the homework and I did the ratio test and it gave the result equal to zero, going off their previous example I assumed that meant it converged only where the series was centered. Well the solutions guide says that the result of the ratio test was zero and therefore converges for all real numbers????!!!!!

chrisplusian · Answer

My laptop is almost dead so I will have to turn it off to charge but if anyone could explain which way it goes, and why I would be greatly appreciative.

anonymous · Answer

An infinite series $\displaystyle \sum a_n$ converges if the ratio test tells you that $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$.

Suppose for example that the given series is $\displaystyle\sum_{n=0}^\infty(x+3)^n$. Since this is a geometric series we know that it will converge for $|x+3|<1$, but let's use the ratio test to determine that:
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{(x+3)^{n+1}}{(x+3)^n}\right|\
~~~~~~~~~~~~~~~~~=\lim_{n\to\infty}\left|x+3\right|\
~~~~~~~~~~~~~~~~~=|x+3|$$
So, for the series to converge, this limit has to be less than 1 (same conclusion). We know the series will diverge if $|x+3|>1$, and we don't know what happens when $|x+3|=1$.

Anyway, solving for $|x+3|<1$ gives you $-1<x+3<1$, or $-4<x<-2$. As you can see, the radius of convergence here is 1.

You also have to test the endpoints. Note that plugging in $x=-4$ or $x=-2$ gives $|x+3|=|\pm1|=1$, so you have the series $\displaystyle\sum_{n=0}^\infty 1^n$, which diverges by the nth term test for divergence.

So, your final interval of convergence was exactly what you found before, $-4<x<-2$, with a radius of 1.

anonymous · Answer

@chrisplusian, but it looks like you're more concerned with series that give you a limit of zero...

Here's an example:
$$\sum_{n=1}^\infty \left(\frac{x+4}{n^2}\right)^n$$
Apply the ratio test:
$$\lim_{n\to\infty}\left|\left(\frac{x+4}{(n+1)^2}\right)^{n+1}\cdot\left(\frac{n^2}{x+4}\right)^n\right|\
|x+4|\lim_{n\to\infty}\frac{1}{(n+1)^2}\left(\frac{n^2}{(n+1)^2}\right)^n\
|x+4|\cdot0=0$$
Since the ratio test says a series converges if the limit of the ratio of successive terms is less than 1, you have that this series converges, since $0<1$.

Because of this, the series converges for any $x$, and so the interval of convergence is $-\infty<x<\infty$, which gives you a radius of $\infty$.

anonymous · Answer

I'm kinda interested in seeing these examples you mentioned.

The first one, you said, gives a limit of infinity as a result of the ratio test. Well, $\infty>1$, which means the series would diverge. Assuming the series was of the form $\displaystyle\sum_{n=0}^\infty (x-c)^n$, the series would only converge if $x=c$, since that would give you $\sum 0$, a convergent series. The interval of convergence would be the point $x=c$, with a radius of 0.

The second one gives a limit of 0, so you would apply the same reasoning as for my second example.

I'm not sure what your last example might be though...

chrisplusian · Answer

Ok I see that I had some of it backwards @SithsAndGiggles but one thing that is still confusing....... a power series has a point at which it converges if the limit equals infinity? I thought that the limit being greater than one implied that the series diverges? How then do we decide that it converges at a single point? This whole power series thing is kind of confusing to me. What is the difference between a power series and a regular series? It seems that we treat them differently. My textbook says only that power series are in the form of.......... and that the can be considered exact values rather than approximations like taylor polynomials. Sorry I know this seems like a great deal to explain but I am going from one college to another. At the college I am at they don't cover this till calculus three and at the college I am about to start attending this was material you are expected to know from calculus two. So I am having to learn this independently and you seem very knowledgable on the subject.

anonymous · Answer

The limit being greater than 1 only means that the series diverges everywhere except the point $c$.

This is because, given some (divergent) power series $\displaystyle\sum_{n=0}^\infty (x-c)^n$, at $x=c$, the series becomes
$$\sum_{n=0}^\infty (c-c)^n=0$$
0 is a finite number, so the sum of a bunch of 0's is also finite. Although the series diverges, it is still convergent for this one point.

Make sense?

anonymous · Answer

The difference between a power series and regular series is that a power series can be considered a type of geometric series.

Something of the form
$$\sum_{n=0}^\infty ar^n$$
would be considered a geometric series.

A power series is a function of $x$, where $x$ is the ratio (replacing $r$):
$$\sum_{n=0}^\infty ax^n$$

chrisplusian · Answer

so the difference is that a power series has a variable for its common ratio?

anonymous · Answer

Right. And applying what you know about regular geometric series, a power series only converges for $|x-c|<1$.

Notice that when $x=c$, you have $|c-c|=0<1$, which would mean any power series converges at $x=c$, like I explained earlier.

chrisplusian · Answer

That is the only thing that has made perfect sense about any of this....thatnks a lot if I could give you a second medal I would.