How to Find the Centre, Radius, and Interval of Convergence for a Power Series?

[danni7070]

[Solved] How to calculate.

Homework Statement

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

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

Homework Equations

Radius of convergence

\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)

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

 

[HallsofIvy]

Homework Statement

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

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

Homework Equations

Radius of convergence

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

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

so R = 1/L

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

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

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 a<sub>n+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)?</sub>

 

[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 don't know what to do with the x in all these calculations.

 

[HallsofIvy]

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.
If the problem were, instead,
\sum_{n=1}^\infty \frac{1}{n}\left(\frac{x+2}{2}\right)^n
or (same thing)
\sum_{n=1}^\infty \frac{1}{n}\frac{(x+2)^n}{2^n}[/itex] <br /> then the radius of convergence would be 2.

 

[danni7070]

You are right! I didn't copy the example right!

It is indeed \sum_{n=1}^\infty \frac{1}{n}\left(\frac{x+2}{2}\right)^n

And then of course the radius of convergence is 2 !

Well, thanks a lot HallsofIvy. I'm kinda happy though that I wasn't doing anything wrong in my calculations, just a "small" typo.

Again thanks.