Video: Using the Limit Comparison Test on a Series

For the series βˆ‘_(𝑛 = 1)^(∞) 1/(𝑒^(2.3𝑛) βˆ’ 4^(𝑛)), use the limit comparison test to determine whether the series converges or diverges.

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Video Transcript

For the series the sum from 𝑛 equals one to ∞ of one divided by 𝑒 to the power of 2.3𝑛 minus four to the 𝑛th power, use the limit comparison test to determine whether the series converges or diverges.

The question gives us a series. It wants us to use the limit comparison test to determine whether our series is convergent or divergent. So let’s start by recalling the limit comparison test. The limit 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 positive sequences for all of our values of 𝑛. Then if the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 divided by 𝑏 𝑛 is some finite, positive constant, we’ll call 𝑐, then either both of our series are convergent or both of our series are divergent.

The best way to use the limit comparison test is to take a look at the series you want to apply it to. Let’s take a closer look at our summand. We can see that the numerator is one. So the only part of our summand which changes as 𝑛 changes is our denominator. So let’s compare the two terms in our denominator, 𝑒 to the power of 2.3𝑛 and four to the 𝑛th power. To compare these, we’ll use our laws of exponents. We’ll rewrite 𝑒 to the power of 2.3𝑛 as 𝑒 to the power of 2.3 all raised to the 𝑛th power. We can then calculate 𝑒 to the power of 2.3. It’s approximately equal to 9.97.

So 𝑒 to the power of 2.3 is approximately 9.97 raised to the 𝑛th power. We know this is bigger than four to the 𝑛th power. So as 𝑛 is approaching ∞, the dominant term in our denominator will be 𝑒 to the power of 2.3𝑛. This means the terms in our series will be getting closer and closer to one divided by 𝑒 to the power of 2.3𝑛. So when we use the limit comparison test, we’ll pick one of our sequences to be one over 𝑒 to the power of 2.3𝑛. In fact, we’ll pick the sequence to be π‘Ž 𝑛. We do this because it will make the denominator in our limit simpler.

So we’re now ready to try and use the limit comparison test. Our sequence π‘Ž 𝑛 is one over 𝑒 to the power of 2.3𝑛. And our sequence 𝑏 𝑛 is the summand of our series one over 𝑒 to the power of 2.3𝑛 minus four to the 𝑛th power. The first thing we need is that our sequence π‘Ž 𝑛 and our sequence 𝑏 𝑛 are positive for all values of 𝑛. In our case, 𝑛 is all of the positive integers. To see this, we’ll start by rewriting our sequence π‘Ž 𝑛 by using our laws of exponents. π‘Ž 𝑛 is 𝑒 to the power of negative 2.3𝑛. And we know 𝑒 raised to the power of any number is always positive. So π‘Ž 𝑛 is positive for all of our values of 𝑛.

Let’s now take a look our sequence 𝑏 𝑛. We see our numerator is positive. So the only way we can have a nonpositive number is if our denominator is negative. However, as we’ve already explained, in our denominator, we have 𝑒 to the power of 2.3𝑛 is bigger than four to the 𝑛th power. We can see this since 𝑒 to the power of 2.3𝑛 is approximately 9.97 raised to the 𝑛th power. So 𝑏𝑛 is also positive for all of our values of 𝑛.

Next, in the limit comparison test, we need to evaluate the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 divided by 𝑏 𝑛. We’ll write this as the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 times the reciprocal of 𝑏 𝑛. Substituting in our expressions for π‘Ž 𝑛 and 𝑏 𝑛 and then simplifying the reciprocal of 𝑏 𝑛, we now have the limit as 𝑛 approaches ∞ of one divided by 𝑒 to the power of 2.3𝑛 multiplied by 𝑒 to the power of 2.3𝑛 minus four to the 𝑛th power.

Next, we’ll distribute one over 𝑒 to the power of 2.3𝑛 over our parentheses. So now we’re dividing each term by 𝑒 to the power of 2.3𝑛. And now we can start simplifying. Our first term is 𝑒 to the power of 2.3𝑛 divided by 𝑒 to the power of 2.3𝑛. This is just equal to one. To simplify our second term, we notice both our numerator and our denominator are raised to the 𝑛th power.

So we’ll use our laws of exponents to write this as four divided by 𝑒 to the power of 2.3 all raised to the 𝑛th power. So this now leaves us with the limit as 𝑛 approaches ∞ of one minus four over 𝑒 to the power of 2.3 all raised to the 𝑛th power. To help us evaluate this limit, we first notice four divided by 𝑒 to the power of 2.3 is approximately 0.4. In particular, this is between negative one and one. So raising this to the 𝑛th power, we get a geometric sequence with ratio of successive terms between negative one and one. And of course, we know a geometric sequence with ratio of successive terms approximately 0.4 will approach zero as 𝑛 approaches ∞. So the second term in our limit approaches zero.

The first term is a constant one. It doesn’t change as the value of 𝑛 changes. So this limit evaluates to give us one. And one is a finite, positive constant. So the second part of our limit comparison test is also true. This means using the limit comparison test, we can conclude either both of our series are convergent or both of our series are divergent. This means we can conclude the convergence or divergence of the series given to us in the question by looking at the convergence or divergence of our sum from 𝑛 equals one to ∞ of π‘Ž 𝑛.

We showed that π‘Ž 𝑛 is 𝑒 to the power of negative 2.3𝑛. We see that this series is a geometric series with our ratio of successive terms π‘Ÿ equal to 𝑒 to the power of negative 2.3. And if we calculate this, we see this is approximately equal to 0.1. And we know all geometric series will be convergent when the absolute value of π‘Ÿ is between negative one and one. So we can conclude our infinite sum of π‘Ž 𝑛 must be convergent. And then, by the limit comparison test, our infinite sum of 𝑏 𝑛 must also be convergent.

Therefore, by using the limit comparison test, we were able to show the sum from 𝑛 equals one to ∞ of one divided by 𝑒 to the power of 2.3𝑛 minus four to the 𝑛th power is convergent.

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