Question Video: Finding the Interval of Convergence of a Power Series | Nagwa Question Video: Finding the Interval of Convergence of a Power Series | Nagwa

# Question Video: Finding the Interval of Convergence of a Power Series Mathematics • Higher Education

Find the interval of convergence for the power series β_(π = 0)^(β) (π₯ β 2)^(π)/(π!).

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### Video Transcript

Find the interval of convergence for the power series the sum from π equals zero to β of π₯ minus two to the πth power over π 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.

Weβre going to use the ratio test to test for convergence of our series. And this says that a series the sum of π π converges if the limit as π approaches β of the absolute value of π π plus one over π π is less than one. In our question, we can say that π π must be equal to π₯ minus two to the πth power over π factorial. And so, π π plus one is π₯ minus two to the power of π plus one over π plus one factorial. Our job is to find the values of π₯ such that the limit as π approaches β 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 π plus one over π plus one factorial times π factorial over π₯ minus two to the πth 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 π plus one divided by π₯ minus two to the power of π is π₯ minus two to the power of one, since π plus one minus π is simply one. We also know that π plus one factorial is π plus one times π times π minus one times π minus two, so on. Which is equal to π plus one times π factorial. And so, we can divide through by π factorial. We now see that the limit as π approaches β of the absolute value of π₯ minus two over π plus one must be less than one. Since π₯ minus two is independent of π, we can take the absolute value of π₯ minus two outside of our limit.

Now, as π approaches β, one over π plus one approaches one over β, 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 π equal zero to β of π₯ minus two to the πth power over π factorial, is convergent for all values of π₯. In this case, we say its interval of convergence is the open interval from negative β to β.

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