Video: Finding the Interval of Convergence of a Power Series

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

02:38

Video Transcript

Find the interval of convergence for the power series the sum from 𝑛 equals zero to ∞ of π‘₯ minus two to the 𝑛th power over 𝑛 factorial.

We recall that the interval of convergence of a power series is the interval that consists of all values of π‘₯ for which the series converges. We can use the ratio or root tests to test for convergence. We do, however, need to be a little bit careful. When π‘₯ is an end point of the interval, the series might converge or diverge at one or both of these end points. And so, the ratio and root tests will always fail when π‘₯ is an end point of the interval in convergence. In this case, we need to check those end points using some other test.

We’re going to use the ratio test to test for convergence of our series. And this says that a series the sum of π‘Ž 𝑛 converges if the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛 is less than one. In our question, we can say that π‘Ž 𝑛 must be equal to π‘₯ minus two to the 𝑛th power over 𝑛 factorial. And so, π‘Ž 𝑛 plus one is π‘₯ minus two to the power of 𝑛 plus one over 𝑛 plus one factorial. Our job is to find the values of π‘₯ such that the limit as 𝑛 approaches ∞ of the absolute value of the quotient of these is less than one. And we know that to divide by a fraction, we multiply by the reciprocal of that fraction. So, our limit becomes π‘₯ minus two to the power of 𝑛 plus one over 𝑛 plus one factorial times 𝑛 factorial over π‘₯ minus two to the 𝑛th power.

We recall that when dividing numbers which have the same base, we can subtract their exponents. And this means π‘₯ minus two to the power of 𝑛 plus one divided by π‘₯ minus two to the power of 𝑛 is π‘₯ minus two to the power of one, since 𝑛 plus one minus 𝑛 is simply one. We also know that 𝑛 plus one factorial is 𝑛 plus one times 𝑛 times 𝑛 minus one times 𝑛 minus two, so on. Which is equal to 𝑛 plus one times 𝑛 factorial. And so, we can divide through by 𝑛 factorial. We now see that the limit as 𝑛 approaches ∞ of the absolute value of π‘₯ minus two over 𝑛 plus one must be less than one. Since π‘₯ minus two is independent of 𝑛, we can take the absolute value of π‘₯ minus two outside of our limit.

Now, as 𝑛 approaches ∞, one over 𝑛 plus one approaches one over ∞, which itself approaches zero. And so, we see that we want to find the values of π‘₯ such that the absolute value of π‘₯ minus two multiplied by zero is less than one. But the absolute value of π‘₯ minus two multiplied by zero will always be zero. And that, in turn of course, is always less than one. And this means our power series, the sum from 𝑛 equal zero to ∞ of π‘₯ minus two to the 𝑛th power over 𝑛 factorial, is convergent for all values of π‘₯. In this case, we say its interval of convergence is the open interval from negative ∞ to ∞.

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