# Question Video: Find the Rational Function Given by a Power Series Mathematics • Higher Education

Consider the power series β_(π = 0)^(β) 3^(π) π₯^(π). Find the function π represented by this series. Determine the interval of convergence of the series.

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### Video Transcript

Consider the power series the sum from π equals zero to β of three to the πth power multiplied by π₯ to the πth power. Find the function π represented by this series. Determine the interval of convergence of this series.

The first thing we want to do is recall a fact about geometric series. Which is that for any constant π, if the absolute value of our ratio π is less than one, then the sum from π equals zero to β of π multiplied by π to the πth power is just equal to π divided by one minus π. And if our constant π is not equal to zero and the absolute value of our ratio π is greater than or equal to one, then the sum from π equals zero to β of π multiplied by π to the πth power does not converge.

To utilize this, we need to rewrite the power series given to us in the question as a geometric series. We notice that our summand three to the πth power multiplied by π₯ to the πth power is just equal to three π₯ all to the πth power. And we see that this gives us the sum from π equals zero to β of three π₯ all to the πth power. Which is a geometric series with an initial term of one and a ratio of three π₯.

So using our rule about geometric series, we must have that this is equal to π divided by one minus π. If the absolute value of π is less than one, we substitute π π is equal to one. And then we substitute π is equal to three π₯. Giving us one divided by one minus three π₯, if the absolute value of three π₯ is less than one. Therefore, we have shown that the function π, which represents this power series, is equal to one divided by one minus three π₯.

To determine the interval of convergence of this series, itβs worth noting that our initial term π is equal to one. So we must have that the absolute value of three π₯ is greater than or equal to one. Then our series does not converge. Therefore, since the series does not converge when the absolute value of three π₯ is greater than or equal to one. We must have that the interval of convergence is what we had before, those values of π₯ which satisfy the absolute value of three π₯ is less than one.

There are several different ways of solving inequalities like this. One such way is if we know the absolute value of three π₯ is less than one. Then we know that three π₯ must be between negative one and one. We can then divide all parts of this inequality by three to get negative a third is less than π₯ is less than one-third. And we know that this is the same as saying that π₯ is in the open interval from negative one-third to a third.

What we have shown is that if access in the open interval from negative a third to a third, then the power series, the sum from π equals zero to β of three to the πth power multiplied by π₯ to the πth power, is equal to the function one divided by one minus three π₯.