Question Video: Using the Comparison Test to Determine the Divergence or Convergence of a Series Mathematics • Higher Education

Use the comparison test to determine whether the series βˆ‘_(𝑛 = 1)^(∞) 1/((1/2)^(𝑛) βˆ’ 3) 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 one-half to the 𝑛th power minus three is convergent or divergent.

The question gives us an infinite series. We want to determine whether this series is convergent or divergent by using the comparison test. Let’s start by recalling the comparison test. 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 both our sequences π‘Ž 𝑛 and 𝑏 𝑛 are greater than or equal to zero for all values of 𝑛, then the comparison test tells us if our sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all of our values of 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 must also converge. And the comparison test also tells us 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 of our values of 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 must also diverge.

The first thing we notice is to use the comparison test, we need both our sequences π‘Ž 𝑛 and 𝑏 𝑛 to be greater than or equal to zero for all of our values of 𝑛. However, this presents us a problem. If we were to substitute 𝑛 is equal to one or in fact any value of 𝑛 into our summand, we would get a negative answer. So at first, it might appear we can’t use the comparison test. Our summand is always negative. However, there’s a trick we can use to still use the comparison test in this case. Since our summand is always negative, if we multiply it by negative one, it will now be positive. Now we just multiply our entire series by negative one. We’ve not changed the convergence or divergence of our series, but we’ve now written every term in our series as a positive number.

So we’ll set π‘Ž 𝑛 to be negative one divided by one-half to the 𝑛th power minus three. This is greater than or equal to zero for all of our values of 𝑛. So we’ve now shown our first prerequisite for the comparison test is true. To continue using the comparison test, we need to decide whether we’re going to show that our series is convergent or divergent. To do this, it will be easier to rewrite our summand. We want to take the negative one in our numerator and move it into our denominator. Doing this and rearranging, we get negative one times the sum from 𝑛 equals one to ∞ of one divided by three minus one-half to the 𝑛th power.

Let’s now take a closer look at the denominator of our summand. We’re subtracting one-half to the 𝑛th power from three. And one-half to the 𝑛th power is positive for all of our values of 𝑛. In fact, this term is making our positive denominator smaller. In other words, we know for all values of 𝑛, three is greater than or equal to three minus one-half to the 𝑛th power. And both of these are positive for all values of 𝑛. This should give us an idea. As 𝑛 is approaching ∞, our one-half to the 𝑛th power term will make less and less of an impact. So we should compare our series to the sum from 𝑛 equals one to ∞ of one over three. In fact, by taking the reciprocal of both sides of our inequality, since both of these are positive, we get one-third is less than or equal to one over three minus one-half to the 𝑛th of power.

And remember, one over three minus one-half to the 𝑛th of power is our term π‘Ž 𝑛. So we’ll try setting 𝑏 𝑛 to be one-third and we’ll want to show that our series is divergent. So let’s start showing our prerequisites for the comparison test. First, we need 𝑏 𝑛 to be greater than or equal to zero for all of our values of 𝑛. Well, 𝑏 𝑛 is a constant one-third for all values of 𝑛. So it’s greater than or equal to zero. Next, we need π‘Ž 𝑛 to be greater than or equal to 𝑏 𝑛 for all of our values of 𝑛. And we’ve already explained why this is true. π‘Ž 𝑛 and 𝑏 𝑛 both have the same numerator. However, 𝑏 𝑛 has a larger, positive denominator. Therefore, 𝑏 𝑛 is smaller than π‘Ž 𝑛.

And of course, this is true for all of our values of 𝑛. So the last thing we need is the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 to be divergent. Well, the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is equal to the sum from 𝑛 equals one to ∞ of one-third. There’s a bunch of different ways to see that this series is divergent. For example, we can take out our factor of one-third and then we see that this is a 𝑝-series, where 𝑝 is equal to zero. However, we can also notice the 𝑛th partial sum is 𝑛 over three. This is an unbounded, partial sum. So the series is divergent.

Therefore, our sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is divergent. And by using the comparison test, we can conclude the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 is divergent. Let’s be extra careful and rigorous with what we’ve just shown. By the comparison test, we’ve shown the sum from 𝑛 equals one to ∞ of one divided by three minus one-half to the 𝑛th power is divergent. But remember, this is not quite equal to the series given to us in the question. We have the same series; however, every term is negative.

But of course we can just write this as multiplying the entire series by negative one. And of course, if this series is divergent, then multiplying the entire series by negative one will also give us a divergent series. Therefore, by using some manipulation and the comparison test, we were able to show the sum from 𝑛 equals one to ∞ of one divided by one-half to the 𝑛th power minus three is a divergent series.

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