Video: Finding the Limit of Rational Functions at Infinity

Consider the series 𝑓(π‘₯) = 1/(1 βˆ’ π‘₯Β²) = βˆ‘_(𝑛 = 0)^(∞) (π‘₯^(2𝑛). Find the interval of convergence for the derivative of the given series.

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Video Transcript

Consider the series 𝑓 of π‘₯ equals one over one minus π‘₯ squared, which equals the sum of π‘₯ to the power of two 𝑛 for values of 𝑛 between zero and ∞. Find the interval of convergence for the derivative of the given series.

We’ve been given a power series expansion for our function. And we’re looking to find the interval of convergence for its derivative. So we begin by recalling that the derivative of the sum of π‘₯ to the power of two 𝑛 for values of 𝑛 between zero and ∞ is equal to the sum of the derivative of π‘₯ to the power of two 𝑛 for values of 𝑛 between zero and ∞.

Now of course when we’re differentiating a polynomial term, we multiply the entire term by its exponent and then reduce the exponent by one. So in this case, the derivative 𝑓 prime of π‘₯ is the sum of two 𝑛 times π‘₯ to the power of two 𝑛 minus one for values of 𝑛 between zero and ∞. So how do we test for convergence?

Well, we recall the ratio test. We’re specifically looking for convergence. So we can use the first part of this test. And it says that if the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛 is less than one, then the series the sum of π‘Ž 𝑛 converges. In our case, π‘Ž 𝑛 is two 𝑛 times π‘₯ to the power of two 𝑛 minus one. So π‘Ž 𝑛 plus one is two times 𝑛 plus one times π‘₯ to the power of two 𝑛 plus one minus one. We divide through by two and distribute our parentheses. And we see that we’re looking for the limit as 𝑛 approaches ∞ of 𝑛 plus one times π‘₯ to the power of two 𝑛 plus one over 𝑛 times π‘₯ to the power of two 𝑛 minus one.

Then we recall that when we divide π‘₯ to the power of two 𝑛 plus one by π‘₯ to the power of two 𝑛 minus one, we subtract their exponents. And this simplifies to π‘₯ squared. π‘₯ is independent of 𝑛. So we can rewrite this as the absolute value of π‘₯ squared times the limit as 𝑛 approaches ∞ of 𝑛 plus one all over 𝑛. We can divide each part of 𝑛 plus one by 𝑛. And then we see as 𝑛 approaches ∞, one plus 𝑛 approaches zero. And this means that the limit as 𝑛 approaches ∞ of one plus one over 𝑛 is simply one. And so we have the absolute value of π‘₯ squared.

Remember, we’re interested in where this converges. So we need to know where this is less than one. So we recall that, for the absolute value of π‘₯ squared to be less than one, the absolute value of π‘₯ itself must be less than one. Which means that π‘₯ must be greater than negative one or less than one. And thus we found the interval of convergence for the derivative of our series. It’s the open interval negative one to one.

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