### 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.

Weβve been given a power series
expansion for our function. And weβre 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 weβre
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. Weβre 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 weβre 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, weβre 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. Itβs the open interval negative one
to one.