Video: Power Series and Radius of Convergence

Find the radius of convergence of the power series βˆ‘_(𝑛 = 0) ^(∞) (𝑛!π‘₯^𝑛)/(2^𝑛).

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Video Transcript

Find the radius of convergence of the power series the sum from 𝑛 equals zero to ∞ of 𝑛 factorial times π‘₯ 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 π‘₯ to the 𝑛th power if the following two statements are true. First, the power series needs to be convergent whenever the absolute value of π‘₯ is less than 𝑅. And the power series needs to be divergent when the absolute value of π‘₯ is greater than 𝑅. And if our power series is convergent for any value of π‘₯, 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 π‘Ž 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 π‘Ž 𝑛 equal to 𝑛 factorial times π‘₯ 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 π‘Ž 𝑛 plus one times the reciprocal of π‘Ž 𝑛. We then substitute in our expression for π‘Ž 𝑛 plus one and π‘Ž 𝑛. Taking the reciprocal gives us the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one factorial times π‘₯ 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 π‘₯ 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 π‘₯ 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 π‘₯ divided by two.

The values of two and π‘₯ 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 π‘₯ divided by two since our limit was of an absolute value. This gives us the absolute value of π‘₯ 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 π‘₯ 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 π‘₯ is less than zero since there are no values of π‘₯ 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 π‘₯ to the 𝑛th power divided by two to the 𝑛th power is zero.

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