Find the radius of convergence and the interval of convergence for: S (n=1 to infinity) (2^n(x-2)^n)/(n+2)!

Question

Find the radius of convergence and the interval of convergence for: Σ (n=1 to infinity) (2^n(x-2)^n)/(n+2)! Nothing really beyond what the question asks. I think I am supposed to find the convergent point and work some magic, but every attempt has me going way off course, so I need a step by step to see where I am going wrong. Thanks. Might be easier to read the question as:

 

Infinity        2^n (x-2)^n

Σ                --------------

n=1                (n+2)!

Answer 1

Use the ratio test. We want

 

lim n-> infinity |a_n+1/a_n| <1

 

where a_n= 2ⁿ(x-2)ⁿ/(n+2)!

 

So we have

 

a_n+1/a_n = 2ⁿ⁺¹ (x-2)ⁿ⁺¹ / (n+3)! * (n+2)!/ ( 2ⁿ (x-2)ⁿ )

 

Note that (n+3)! shows up because we have ((n+1)+2)! = (n+1+2)!= (n+3)!

 

Now we need to simply, I'll group the terms a bit (though it's hard to type on here)

 

2ⁿ⁺¹/2ⁿ  * (x-2)ⁿ⁺¹/(x-2)ⁿ   * (n+2)!/(n+3)!

 

Now observe that 

2ⁿ⁺¹=2ⁿ2

and

(n+3)!=(n+3)(n+2)(n+1)n!

(n+2)!=(n+2)(n+1)n!

 

So that we can cancel a bit, I'll trust you to do that to get

 

2 * (x-2) * 1/(n+3)   = (x-2) * 2/(n+3)

 

Now we take the limit as n tends to infinity. But notice as n gets very large, 2/(n+3) gets very small (the limit is 0) so we have

 

lim n to infinity |a_n+1/a_n|= lim n to infinity |(x-2)*2/(n+3) |= |x-2| * lim n to infinity |2/(n+3)| = |x-2| * 0 =0

 

We need to check for which x this is <1. But notice the limit is 0 no matter what x we choose and 0<1 always! So we can choose any x so that the radius of convergence is infinite. To see an example where this was not zero, suppose the limit was 1. Then we would want

 

|x-2|* 1<1

|x-2|<1

So that the radius of convergence is 1 and the interval of convergence would be

-1 < x-2 < 1

1 < x < 3

Then just check the end points; that is, the cases where x=1 and x=3 and check for convergence there. Suppose the series did not converge for x=1 but did converge for x=3, then we would have the interval

1 < x <= 3