What is a power series that converges on (3,6) but not on any larger interval?

[sassie]

Homework Statement

Give an example of a power series tha converges on the interval (3,6), but on no larger interval. Give some justification.

Homework Equations

The equation needed is probably that for a power series:

$$\Sigma$$ c_n(x-a)ⁿ

The Attempt at a Solution

I'm not sure at all. The ratio test maybe, but I don't know where to start. Any pointers would be great.

 

[Bacat]

The power series converges inside the radius of convergence. You should know how to solve a power series to find it's radius of convergence. Write this out on a piece of paper and look at it. When you understand which part is the radius (interval) of convergence, you can plug in the numbers you need. Then rearrange the equation back into power series form and you should be done.

So, first solve the power series for the radius generally (with variables, not numbers). Then plug in the numbers you need for the radius of convergence. Then unsolve the power series using those numbers.

 

[sassie]

So (please correct me if I'm wrong) we need to use the formula for the power series, as I have given in the problem. We then perform a test on it (ratio, root etc.), and find to find, as you said, the general solution to the power series.

So, using the ratio test, we get |(a(n+1)/a(n)|=c1(x-a)

The "c1(x-a)" competent is the bit that needs to be less than 1 (i.e. it is the radius of convergence) so that the series converges. But from then on I don't exactly know what to do. I can't see where I can sub the numbers in (because of the difference between the endpoints of the interval is 3...).

 

[zcd]

A good series to use for small intervals of convergence is a geometric series.
using ratio test:
$$\sum_{n=0}^{\infty} x^{n}$$ converges on $$\lim_{n\to\infty}\left|\frac{x^{n+1}}{x^{n}}\right| \implies |x|<1$$
$$\sum_{n=0}^{\infty} (x-a)^{n}$$ converges on $$\lim_{n\to\infty}\left|\frac{(x-a)^{n+1}}{(x-a)^{n}}\right| \implies |x-a|<1$$, and x converges on a-1<x<a+1
$$\sum_{n=0}^{\infty} (bx)^{n}$$ converges on $$\lim_{n\to\infty}\left|\frac{(bx)^{n+1}}{(bx)^{n}}\right| \implies |x|<b$$

 

[sassie]

A good series to use for small intervals of convergence is a geometric series.
using ratio test:
$$\sum_{n=0}^{\infty} x^{n}$$ converges on $$\lim_{n\to\infty}\left|\frac{x^{n+1}}{x^{n}}\right| \implies |x|<1$$
$$\sum_{n=0}^{\infty} (x-a)^{n}$$ converges on $$\lim_{n\to\infty}\left|\frac{(x-a)^{n+1}}{(x-a)^{n}}\right| \implies |x-a|<1$$, and x converges on a-1<x<a+1
$$\sum_{n=0}^{\infty} (bx)^{n}$$ converges on $$\lim_{n\to\infty}\left|\frac{(bx)^{n+1}}{(bx)^{n}}\right| \implies |x|<b$$

So all I need to do is to pick one of these and show that the power series in one of these forms converges? And, let's say I choose $$\sum_{n=0}^{\infty} (x-5)^{n}$$ and have an interval of (4,6). Because it the question requires it to converge on (3,6), could I pick this given series?

 

[sassie]

ha ha ha, I've just figured it out. thanks for all your help!