# How to calculate.

1. Nov 23, 2007

### danni7070

[Solved] How to calculate.

1. The problem statement, all variables and given/known data

Determine the centre, radius, and Interval of convergence of the power series.

$$\sum_{n=1}^\infty \frac{1}{n} \frac{(x+2)}{2}^n$$

2. Relevant equations

$$\frac{1}{R} = L = \lim_{n \to {\infty}} \frac{a_n+1}{a_n}$$

so R = 1/L

and the interval is (x-R,x+R)

3. The attempt at a solution

Finding the centre is easy. x = -2

But the algebra stuff is getting me down. This is where I get confused. I took this from the solution manual and I can't figure out this result.

$$\sum_{n=1}^\infty \frac{1}{n} \frac{(x+2)}{2}^n$$

So we have $$R = \lim \frac{2^{n+1}(n+1)}{2^n n} = 2$$

Last edited: Nov 23, 2007
2. Nov 23, 2007

### HallsofIvy

Staff Emeritus
Put { } around the "n+1" subscript.

Did you copy both the original problem and the solution correctly? In your original formula, that "2" in the denominator was NOT to the n power and so you will NOT have 2n in your ratio. In fact the factors of "2" in both an+1[/sup] and an will cancel and play no part. It should be obvious that
$$\sum_{n=1}^\infty \frac{1}{n}\frac{(x+2)^n}{2}$$
is the same as
$$\frac{1}{2}\sum_{n=1}^\infty \frac{1}{n}\(x+2)^n$$
so the "1/2" plays no part in the convergence of the series.

You are given that $a_n= (1/n)(x+2)^n$ so $a_{n+1}= (1/(n+1))(x+2)^{n+1}$. According to the "ratio test", that will converge as long as, in the limit, the ratio of consectutive terms goes to a limit of less than 1. The ratio you want is
$$\frac{1}{n+1}\frac{|x+2|^{n+1}}{2}\frac{n}{1}\frac{2}{|x+2|^n}$$
which reduces to
$$\frac{n}{n+1}|x+2|}$$
What is the limit of n/(n+1)?

3. Nov 23, 2007

### danni7070

Ok, first of all, thanks for taking time looking at this.

Limit of n/(n+1) is 1

This is exactly what I got. But how is R = 2 then?

Is it right to say that the limit of $$\frac{n}{n+1}|x+2|}$$ is 2 ?

I dont know what to do with the x in all these calculations.

4. Nov 23, 2007

### HallsofIvy

Staff Emeritus
Now, the limit of $$\frac{n}{n+1}|x+2|$$ is |x+2|. That will be less than 1 as long as -1< x+2< 1 or =-3< x< -1. The radius of convergence is 1.

That was why I asked if you were sure you had copied everything correctly.
$$\sum_{n=1}^\infty \frac{1}{n}\left(\frac{x+2}{2}\right)^n$$