### Video Transcript

Find the radius of convergence of

the power series the sum from 𝑛 equals zero to ∞ of 𝑛 factorial times 𝑥 to the

𝑛th power all divided by two to the 𝑛th power.

The question wants us to find the

radius of convergence of this power series. And we call 𝑅 the radius of

convergence of the power series the sum of 𝑐 𝑛 𝑥 to the 𝑛th power if the

following two statements are true. First, the power series needs to be

convergent whenever the absolute value of 𝑥 is less than 𝑅. And the power series needs to be

divergent when the absolute value of 𝑥 is greater than 𝑅. And if our power series is

convergent for any value of 𝑥, then we say that our radius of convergence 𝑅 is

positive ∞.

To help us find the radius of

convergence of the power series given to us in the question, we’re going to use the

ratio test. And we recall the ratio test tells

us if we have an infinite series the sum of 𝑎 𝑛, where the limit as 𝑛 approaches

∞ of the absolute value of the ratio of successive terms is less than one, then we

can conclude that our series is absolutely convergent. However, if the limit as 𝑛

approaches ∞ of the absolute value of the ratio of successive terms is greater than

one, then our series must be divergent. So to apply the ratio test on our

series, we’ll set 𝑎 𝑛 equal to 𝑛 factorial times 𝑥 to the 𝑛th power divided by

two to the 𝑛th power.

To start, we notice the limit as 𝑛

approaches ∞ of the absolute value of the ratio of successive terms is the same as

saying the limit as 𝑛 approaches ∞ of the absolute value of 𝑎 𝑛 plus one times

the reciprocal of 𝑎 𝑛. We then substitute in our

expression for 𝑎 𝑛 plus one and 𝑎 𝑛. Taking the reciprocal gives us the

limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one factorial times 𝑥 to

the power of 𝑛 plus one divided by two to the power of 𝑛 plus one times two to the

𝑛th power divided by 𝑛 factorial times 𝑥 to the 𝑛th power.

We can simplify this expression by

cancelling 𝑛 of our shared factors of two in the numerator and the denominator and

𝑛 of the shared factors of 𝑥 in the numerator and the denominator. We also have that 𝑛 plus one

factorial is equal to 𝑛 plus one multiplied by 𝑛 factorial. So we can cancel the shared factor

of 𝑛 factorial in our numerator and our denominator. So we’ve simplified our limit to be

the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one times 𝑥 divided

by two.

The values of two and 𝑥 don’t

change as our limit of 𝑛 is approaching ∞. So these factors aren’t variant

with respect to 𝑛. So we can take these outside of our

limit. Well, we must remember to use the

absolute value of 𝑥 divided by two since our limit was of an absolute value. This gives us the absolute value of

𝑥 over two multiplied by the limit as 𝑛 approaches ∞ of the absolute value of 𝑛

plus one. And we see as 𝑛 is approaching ∞,

one remains constant however 𝑛 is approaching ∞. So 𝑛 plus one is approaching

∞. And the absolute value of 𝑛 plus

one is approaching ∞.

So we’ve shown that this is equal

to some constant multiplied by a limit which is approaching ∞. So the only way that this limit can

be less than one is if our constant was equal to zero. So we set our radius of convergence

equal to zero. Then, by the ratio test, we’ve

shown that our series must be divergent when the absolute value of 𝑥 is greater

than zero because the limit as 𝑛 approaches ∞ of the absolute value of the ratio of

successive terms will be greater than one. And it’s always true that the

series will be convergent when the absolute value of 𝑥 is less than zero since

there are no values of 𝑥 who have an absolute value less than zero.

Therefore, we’ve shown the radius

of convergence for the power series the sum from 𝑛 equals zero to ∞ of 𝑛 factorial

times 𝑥 to the 𝑛th power divided by two to the 𝑛th power is zero.