Finding the radius of convergence

[Timebomb3750]

Finding the radius of convergence...

Homework Statement

1+2x+(4x^(2)/2!)+(8x^(3)/3!)+(16x^(4)/4!)+(32x^(5)/5!)+...

Homework Equations

I would use the ratio test. Which is...

lim as n→∞ (An+1/An)

The Attempt at a Solution

I know what to do to find the answer, but I don't know how to find the expression for the provided series.

I'm guessing it's ((2x^(n))/n!), then work with the ratio test from there?

 

[Char. Limit]

First, I assume you mean ((2x)^n)/n!. That IS correct. Use the ratio test from here. Remember that the ratio has to be less than 1 (as n goes to infinity) for the series to converge. That should help you out.

 

[Timebomb3750]

First, I assume you mean ((2x)^n)/n!. That IS correct. Use the ratio test from here. Remember that the ratio has to be less than 1 (as n goes to infinity) for the series to converge. That should help you out.

Sweet. I was right on the expression. Okay, so I did the ratio test.

(2x^(n+1))/((n+1)!) * (n!/2x^(n)) = ((xn!)/(n+1)!) But doesn't this lim as n→∞ of this equal 0? Or is there more simplification I need to do? The answer is (1/4), but I don't see how I'm going to get that.

 

[Char. Limit]

First, the ratio test gives zero, i.e. less than 1 for any x you choose, so what does that tell you about the radius?

1/4 is definitely not the correct answer for this function. I checked that two different ways, and they gave me the same answer... and it wasn't 1/4.

 

[Timebomb3750]

First, the ratio test gives zero, i.e. less than 1 for any x you choose, so what does that tell you about the radius?

1/4 is definitely not the correct answer for this function. I checked that two different ways, and they gave me the same answer... and it wasn't 1/4.

Odd. Because the textbook says the answer to this is (1/4). Well. I should double check to see if my expression agrees with the given series.

 

[Timebomb3750]

Oops! Wow. I was doing the wrong problem.

The series is 1+2x+((4!x^(2))/(2!))+((6!x^(3))/((3!^(2))+...

Sorry. Okay. I know the denominator of the expression is going to be (n!)^2, but not so sure of the numerator. Other than that, I probably can handle the rest of the problem.

 

[Timebomb3750]

Oops! Wow. I was doing the wrong problem.

The series is 1+2x+((4!x^(2))/(2!))+((6!x^(3))/((3!^(2))+...

Sorry. Okay. I know the denominator of the expression is going to be (n!)^2, but not so sure of the numerator. Other than that, I probably can handle the rest of the problem.

How do I edit posts? I keep typing the series wrong...

 

[Timebomb3750]

Forget it. Delete this. I'll just make a new thread asking for help on finding the series. I'll be sure to type out the series correctly this time. Sorry. I feel so stupid.

 

[Char. Limit]

No need to worry I figured out what you meant. The denominator is (n!)^2, like you said. The numerator isn't that hard, either, actually. Just think about what sequence goes 0!, 2!, 4!, 6!.