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Find the radius of convergence and interval of co…

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Problem 12 Easy Difficulty

Find the radius of convergence and interval of convergence of the series.
$$\sum_{n=1}^{\infty} \frac{(2 n) !}{2^{n}} x^{n}$$

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Calculus 2 / BC

Calculus

Chapter 8

Infinite Sequences and Series

Section 5

Power Series

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Video Transcript

in this question, we're asked to find the radius of convergence and the interval of convergence of the following series. To do that. We're going to use the ratio test by the ratio test we first need to calculate the limit of absolute value of A. M plus one over a M as N goes to infinity. Here I am is a general term of the series. In our case this is going to be limited absolute value of now we need to replace and by N plus one everywhere We're going to get to and plus two material Times X to the N Plus one divided by two to the N plus one, multiplied by the reciprocal of am which is going to be two to the end over to in fact Toral times X to the end. This is equal to the limit absolute reality of X times one half Now, 20 plus two factorial Over two. In factorial gives us 20 plus one Times two and Plus two. And since x here is a fixed number and n goes to infinity 20 plus one times 20 plus two also goes to infinity and infinity times a fixed number is going to be infinity. By the ratio test for the series to converge. We want this limit to be less than one and if this limit is greater than one, this means that the series diverges. Now note that this limit is always equal to equals to infinity independently or the value of X meaning this limit is always always greater than one for all values of X. And this means that our series diverges for all values of X, except for X equals zero, so it diverges for all X non zero. This means that the radius of convergence is equal to zero Because our series converges only at one point, and the interval of convergence is just one point, it's just X equals zero, that's all.

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