Video: Determining Whether a Given Series Is Convergent or Divergent Using the Comparison Test

Use the comparison test to determine whether the series βˆ‘_(𝑛 = 1) ^(∞) (1)/(3^(𝑛) + 5) is convergent or divergent.

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Video Transcript

Use the comparison test to determine whether the series the sum from 𝑛 equals one to ∞ of one divided by three to the 𝑛th power plus five is convergent or divergent.

The question wants us to determine the convergence or divergence of the series by using the comparison test. To use the comparison test, we need a series with which we will compare the series given to us in the question. And if we look at the series given to us in the question, the only part of our summand which changes as 𝑛 changes is the term three to the 𝑛th power in our denominator. So it would make sense that our series will behave similar to the sum from 𝑛 equals one to ∞ of one divided by three to the 𝑛th power.

And we recall the comparison test tells us if we have two series the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 where π‘Ž 𝑛 and 𝑏 𝑛 are both greater than or equal to zero for all integers 𝑛 greater than or equal to one. Then the comparison test tells us if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges, and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all integers 𝑛 greater than or equal to one. Then we can conclude the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 must also converge.

And we also have if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 diverges, and π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all integers 𝑛 greater than or equal to one, then we can conclude the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 must diverge. And we want to compare the series given to us in the question with the sum from 𝑛 equals one to ∞ of one divided by three to the 𝑛th power. This is a geometric series with ratio of successive terms π‘Ÿ equal to one-third.

And we know that this is convergent since any geometric series whose absolute value of ratio of successive terms π‘Ÿ is less than one will converge. Since this series converges, we’ll try to use the first part of the comparison test. We’ll set π‘Ž 𝑛 to be the summand of the series given to us in the question. That’s one divided by three to the 𝑛th power plus five. And we’ll set 𝑏 𝑛 to be one divided by three to the 𝑛th power.

First, to use the comparison test, we need to show that both π‘Ž 𝑛 and 𝑏 𝑛 are greater than or equal to zero for all of our possible values of 𝑛. And we know that three to the 𝑛th power is greater than zero for any value of 𝑛. This tells us that our denominator of π‘Ž 𝑛, three to the 𝑛th power plus five, is positive and our denominator of 𝑏 𝑛, which is three to the 𝑛th power, is also positive. And one divided by a positive number is positive. So both π‘Ž 𝑛 and 𝑏 𝑛 are greater than or equal to zero for all integers 𝑛 greater than or equal to one.

Next, we need to show that the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges. And we’ve already done this. We chose this series to be a geometric series whose ratio of successive terms π‘Ÿ is equal to one-third. And since the absolute value of this is less than one, we know that this has to converge. The last thing we need to show for the comparison test is that π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all integers 𝑛 greater than or equal to one. To help us see why this is true, let’s compare the denominators of our two sequences.

We know that adding five makes a number bigger. So three to the 𝑛th power plus five must be bigger than three to the 𝑛th power. And we know that both of these are always positive. So what does this tell us? Well, π‘Ž 𝑛 always has the larger positive denominator. So when we’re calculating π‘Ž 𝑛, we’re dividing one by a larger positive number. So π‘Ž 𝑛 must be smaller than 𝑏 𝑛. So we’ve shown that π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all of our possible values of 𝑛.

So we’ve now shown all the prerequisites for the comparison test are true. So we can conclude using the comparison test that the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 converges. In conclusion, we’ve shown by using the comparison test that the sum from 𝑛 equals one to ∞ of one divided by three to the 𝑛th power plus five is convergent.

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