# Video: Determining Whether a Given Series Is Absolutely Convergent, Conditionally Convergent, or Divergent

Determine if the series β_(π = 1) ^(β) (β2)^(π) 3^(1 β π) is absolutely convergent, conditionally convergent, or divergent.

04:19

### Video Transcript

Determine if the series the sum from π equals one to β of negative two to the πth power multiplied by three to the power of one minus π is absolutely convergent, conditionally convergent, or divergent.

The question gives us a series and wants to determine if this series is absolutely convergent, conditionally convergent, or divergent. And we recall we call the series the sum from π equals one to β of π π absolutely convergent if the sum from π equals one to β of the absolute value of π π is convergent. We call the series conditionally convergent if the sum from π equals one to β of π π converges and the series is not absolutely convergent. Otherwise, we say that our series is divergent.

Weβll start by checking that our series is absolutely convergent. So we need to check that the sum from π equals one to β of the absolute value of our summand is convergent. Taking the absolute value of the summand of the series given to us in the question gives us the sum from π equals one to β of the absolute value of negative two to the πth power multiplied by three to the power of one minus π. We can simplify our summand by using our laws for exponents. We have that negative two is negative one times two. So negative two all raised to the πth power is equal to negative one to the πth power multiplied by two to the πth power. This gives us the sum from π equals one to β of the absolute value of negative one to the πth power times two to the πth power times three to the power of one minus π.

Next, weβre going to simplify our series by using a fact about the absolute value. We have that the absolute value of a product of terms is equal to the product of their absolute values. We see that negative one raised to the πth power will always give us either negative one or one depending on whether π is odd or even. However, the absolute value of both of these is equal to one. So the absolute value of negative one to the πth power will always give us one. And we can see that two raised to the πth power is always positive and three to the power of one minus π is also always positive. And applying the absolute value to a positive number doesnβt change anything. This gives us the sum from π equals one to β of two to the πth power times three to the power of one minus π.

We can again simplify our summand by using our laws for exponents. We have that three to the power of one minus π is just equal to three times three to the power of negative π. Next, weβre going to rewrite three to the power of negative π as one divided by three to the πth power. This gives us the sum from π equals one to β of two to the πth power times three divided by three to the πth power. We can again simplify this by using more of our laws for exponents. We have that π to the πth power divided by π to the πth power is equal to π over π all raised to the πth power. Using this, we can rewrite two to the πth power divided by three to the πth power as two-thirds all raised to the πth power.

We can now see that our sum is almost in the form of a geometric series. And we also know that the geometric series the sum from π equals one to β of π times π to the power of π minus one will converge if the absolute value of π is less than one. Instead of two-thirds raised to the πth power, we want two-thirds raised to the power of π minus one. And we can do this by writing two-thirds raised to the πth power as two-thirds to the power of π minus one multiplied by two-thirds. We can then simplify this by canceling the shared factor of three in our numerator and our denominator. This gives us that our sum is a geometric series with the ratio of successive terms π equal to two-thirds and initial value π equal to two. And we know a geometric series will converge if the absolute value of π is less than one.

So when π π is the summand of the series given to us in the question, weβve shown that the sum from π equals one to β of the absolute value of π π is convergent. Which means weβve shown the series given to us in the question is absolutely convergent. And since any series can only be one of absolutely convergent, conditionally convergent, or divergent, we donβt need to check anymore; weβre done. Therefore, weβve shown that the sum from π equals one to β of negative two to the πth power times three to the power of one minus π is an absolutely convergent series.