Radius/Interval Convergence for Power Series

[mcdowellmg]

Homework Statement

Find the radius and interval of convergence for the power series of n=0 of infinity of n^3(x-5)^n

Homework Equations

Ratio test: http://en.wikipedia.org/wiki/Ratio_test

The Attempt at a Solution

[(n+1)^3(x-5)^n+1 / n^3(x-5)^n]

I am lost as to how to simplify this solution to find out what the limit of n approaching infinity would be (for the radius of convergence). Since I can't find out the radius, I can't find out the interval. Any tips? I also have one other similar problem that I can't figure out for the same reason. Thanks for any help!

 

[HallsofIvy]

It's fairly simple to reduce that to
$$\frac{(n+1)^3}{n^3}(x- 5)= \frac{n^3+ 3n^2+ 3n+ 1}{n^3}(x- 5)= (1+ \frac{3}{n}+ \frac{3}{n^2}+ \frac{1}{n^3})(x- 5)$$
It should be easy to then take the limit as n goes to infinity.

 

[mcdowellmg]

Thanks! I should have noticed that one. I found the limit to be 1 and then the interval to be 5-R < x < 5+R, so 4 < x < 6.

 

[HallsofIvy]

You can extend that slightly. Notice that at x= 4, x- 5= -1 so the sum becomes
$$\sum \frac{(-1)^n}{n^3}$$
which obviously converges because it is an alternating series with absolute value going to 0.

At x= 6, x- 5= 1 so the sum becomes
$$\sum \frac{1}{n^3}$$
which is a "p-series" with p= 3 and so converges. The interval of convergence is actually
$$4\le x\le 6$$.

 

[mcdowellmg]

Thanks! I have one more problem, if you don't mind.

$$\sum$$ of n=1 to infinity of $$\frac{(2)^{n}}{\sqrt{n}}$$ * (x+3)$$^{n}$$

I tried to simplify it down to 1/2 by cancelling out the square roots of n and also the (-2)^{n} and (-2)^{n+1} and the (x+3)^{n} and (x+3)^{n+1} to 1 + 1 = 2. Perhaps I am going at that the wrong way?

I also thought that maybe the n=1 would make it an integral of a power series? That might be a way I should look at the problem.

 

[Mark44]

For future reference, here's what your series looks like in LaTeX:
$$\sum_{n = 1}^{\infty}\frac{2^n~(x + 3)^n}{\sqrt{n}}$$

Form the ratio a_n+1/a_n and take the limit. You shouldn't be getting 1/2.

Also, why do you have (-2)ⁿ? You don't show this in what you posted.

 

[mcdowellmg]

Sorry, the latex was tricky to figure out, and in the confusion I left off the - in -2 to the nth power.

 

[Mark44]

I thought it was something like that. For the LaTeX, click the summation expression in my last post to see how to do it.

 

[HallsofIvy]

This is almost exactly the same as the previous problem- where you said "I should have noticed that one"!
So
[tex]a_n= \frac{(-1)^n}{\sqrt{n}}(x+ 3)^n[/itex]<br /> then<br /> $$a_{n+1}= \frac{(-1)^{n+1}}{\sqrt{n+1}}(x+3)^{n+1}$$<br /> <br /> Since the ratio test applies to positive series, we take the absolute values so we can ignore the powers of -1. <br /> <br /> $$\frac{a_{n+1}}{a_n}= \frac{1}{\sqrt{n+1}}\frac{\sqrt{n}}{1}\frac{|x+ 3|^{n+1}}{|x+ 1|^n}$$<br /> $$= \sqrt{\frac{n+1}{n}}|x+3|$$<br /> Again, divide both numerator and denominator of the fraction by the highest power of n, here n:<br /> $$= \sqrt{\frac{1+ 1/n}{1}}|x+3|$$[/tex]