Question Video: Finding the Limit of Rational Functions at Infinity | Nagwa Question Video: Finding the Limit of Rational Functions at Infinity | Nagwa

# Question Video: Finding the Limit of Rational Functions at Infinity Mathematics • Higher Education

Consider the series π(π₯) = 1/(1 β π₯Β²) = β_(π = 0)^(β) (π₯^(2π). Find the interval of convergence for the derivative of the given series.

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### Video Transcript

Consider the series π of π₯ equals one over one minus π₯ squared, which equals the sum of π₯ to the power of two π for values of π between zero and β. Find the interval of convergence for the derivative of the given series.

Weβve been given a power series expansion for our function. And weβ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 π for values of π between zero and β is equal to the sum of the derivative of π₯ to the power of two π for values of π between zero and β.

Now of course when weβ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 π prime of π₯ is the sum of two π times π₯ to the power of two π minus one for values of π between zero and β. So how do we test for convergence?

Well, we recall the ratio test. Weβre specifically looking for convergence. So we can use the first part of this test. And it says that if the limit as π approaches β of the absolute value of π π plus one over π π is less than one, then the series the sum of π π converges. In our case, π π is two π times π₯ to the power of two π minus one. So π π plus one is two times π plus one times π₯ to the power of two π plus one minus one. We divide through by two and distribute our parentheses. And we see that weβre looking for the limit as π approaches β of π plus one times π₯ to the power of two π plus one over π times π₯ to the power of two π minus one.

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

Remember, weβ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. Itβs the open interval negative one to one.

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