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

Determine if the series βˆ‘_(𝑛 = 1) ^(∞) (βˆ’2)^(𝑛) 3^(1 βˆ’ 𝑛) is absolutely convergent, conditionally convergent, or divergent.

04:19

Video Transcript

Determine if the series the sum from 𝑛 equals one to ∞ of negative two to the 𝑛th power multiplied by three to the power of one minus 𝑛 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 𝑛 equals one to ∞ of π‘Ž 𝑛 absolutely convergent if the sum from 𝑛 equals one to ∞ of the absolute value of π‘Ž 𝑛 is convergent. We call the series conditionally convergent if the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 converges and the series is not absolutely convergent. Otherwise, we say that our series is divergent.

We’ll start by checking that our series is absolutely convergent. So we need to check that the sum from 𝑛 equals one to ∞ 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 𝑛 equals one to ∞ of the absolute value of negative two to the 𝑛th power multiplied by three to the power of one minus 𝑛. 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 𝑛th power is equal to negative one to the 𝑛th power multiplied by two to the 𝑛th power. This gives us the sum from 𝑛 equals one to ∞ of the absolute value of negative one to the 𝑛th power times two to the 𝑛th power times three to the power of one minus 𝑛.

Next, we’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 𝑛th power will always give us either negative one or one depending on whether 𝑛 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 𝑛th power will always give us one. And we can see that two raised to the 𝑛th power is always positive and three to the power of one minus 𝑛 is also always positive. And applying the absolute value to a positive number doesn’t change anything. This gives us the sum from 𝑛 equals one to ∞ of two to the 𝑛th power times three to the power of one minus 𝑛.

We can again simplify our summand by using our laws for exponents. We have that three to the power of one minus 𝑛 is just equal to three times three to the power of negative 𝑛. Next, we’re going to rewrite three to the power of negative 𝑛 as one divided by three to the 𝑛th power. This gives us the sum from 𝑛 equals one to ∞ of two to the 𝑛th power times three divided by three to the 𝑛th power. We can again simplify this by using more of our laws for exponents. We have that π‘Ž to the 𝑛th power divided by 𝑏 to the 𝑛th power is equal to π‘Ž over 𝑏 all raised to the 𝑛th power. Using this, we can rewrite two to the 𝑛th power divided by three to the 𝑛th power as two-thirds all raised to the 𝑛th 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 𝑛 equals one to ∞ of π‘Ž times π‘Ÿ to the power of 𝑛 minus one will converge if the absolute value of π‘Ÿ is less than one. Instead of two-thirds raised to the 𝑛th power, we want two-thirds raised to the power of 𝑛 minus one. And we can do this by writing two-thirds raised to the 𝑛th power as two-thirds to the power of 𝑛 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‘Ÿ equal to two-thirds and initial value π‘Ž equal to two. And we know a geometric series will converge if the absolute value of π‘Ÿ is less than one.

So when π‘Ž 𝑛 is the summand of the series given to us in the question, we’ve shown that the sum from 𝑛 equals one to ∞ of the absolute value of π‘Ž 𝑛 is convergent. Which means we’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 don’t need to check anymore; we’re done. Therefore, we’ve shown that the sum from 𝑛 equals one to ∞ of negative two to the 𝑛th power times three to the power of one minus 𝑛 is an absolutely convergent series.

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