Video: Use the Limit Comparison Test to Determine the Convergence or Divergence of a Series

For the series βˆ‘_(𝑛 = 1)^(∞) 1/(5^𝑛 βˆ’ 2^𝑛), use the limit comparison test to determine whether the series converges or diverges.

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Video Transcript

For the series which is equal to the sum from 𝑛 equals one to ∞ of one divided by five to the 𝑛th power minus two to the 𝑛th power, use the limit comparison test to determine whether the series converges or diverges.

The question tells us to use the limit comparison test. So let’s recall that the limit comparison test tells us that if we have a sequence π‘Ž 𝑛 which is greater than or equal to zero and a sequence 𝑏 𝑛 which is greater than zero, where 𝑛 is greater than or equal to one. And that the limit as 𝑛 approaches ∞ of the quotient π‘Ž 𝑛 over 𝑏 𝑛 is equal to 𝑐, where 𝑐 is a finite positive number. Then we can conclude that the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 both converge or both diverge.

The concluding statement in the limit comparison test tells us that the convergence and divergence of these two series are the same. So if we can find sequences π‘Ž 𝑛 and 𝑏 𝑛 such that the prerequisites of the limit comparison test are true and where we can determine where one of the series either converges or diverges. Then we can use the limit comparison test to determine whether the other series converges or diverges.

Since the question is asking us about the sum from 𝑛 equals one to ∞ of one divided by five to the 𝑛th power minus two to the 𝑛th power. By using the limit comparison test, we should set however π‘Ž 𝑛 or 𝑏 𝑛 to be equal to one divided by five to the 𝑛th power minus two to the 𝑛th power. Let’s try setting this equal to be 𝑛. We will see why shortly. Now, let’s see what expression we get for our limit of our quotient π‘Ž 𝑛 over 𝑏 𝑛, where we have that this limit is equal to the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 multiplied by the reciprocal of 𝑏 𝑛. And we know the reciprocal of 𝑏 𝑛 is just five to the 𝑛th power minus two to the 𝑛th power. This gives us that the limit of our quotient is equal to the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 multiplied by five to the 𝑛th power minus two to the 𝑛th power. And we want this to be equal to 𝑐, where 𝑐 is some finite positive number.

We see if we just take the limit of five to the 𝑛th power minus two to the 𝑛th power, then this gets bigger and bigger as 𝑛 gets bigger and bigger. Because five is greater than two, so this limit is to ∞. Since we want the limit of this product to be some finite positive number, we can conclude that the limit of π‘Ž 𝑛 as 𝑛 approaches ∞ must be equal to zero. At this point, let’s let π‘Ž 𝑛 be equal to one divided by 𝑑 𝑛 and substitute this into our limit. This gives us that we want the limit as 𝑛 approaches ∞ of five to the 𝑛th power minus two to the 𝑛th power divided by our new sequence 𝑑 𝑛 to be equal to some finite positive number.

One thing we could try is setting our sequence 𝑑 𝑛 to be equal to five to the 𝑛th power because five to the 𝑛th power is the fastest growing part of our numerator. We can then split this fraction so that we’re dividing both terms in the numerator by five to the 𝑛th power. And we know that five to the 𝑛th power divided by five to the 𝑛th power is just equal to one. Then we can use the fact that π‘₯ to the 𝑛th power divided by 𝑦 to the 𝑛th power is equal to π‘₯ over 𝑦 all raised to the 𝑛th power. To see that two to the 𝑛th power divided by five to the 𝑛th power is just two-fifths raised to the 𝑛th power. This gives us the limit as 𝑛 approaches ∞ of one minus two-fifths to the 𝑛th power.

Now, we know that the limit as 𝑛 approach ∞ of π‘Ÿ to the 𝑛th power is equal to zero when the size of π‘Ÿ is less than one. So we can use this to conclude that the limit of two-fifths to the 𝑛th power as 𝑛 approach ∞ is equal to zero. Therefore, since the limit of one as 𝑛 approaches ∞ is just equal to one, we have that the limit of π‘Ž 𝑛 over 𝑏 𝑛 as 𝑛 approaches ∞ is just equal to one. Since one is a finite positive number, we have shown one of the prerequisites for our limit comparison test.

Next, let’s check if the sequence π‘Ž 𝑛 is nonnegative for 𝑛 greater than or equal to one. We have that the sequence π‘Ž 𝑛 is equal to one over five to the 𝑛th power which we can rewrite as one-fifth to the 𝑛th power. Since 𝑛 is a positive integer, one-fifth to the 𝑛th power is just one-fifth multiplied by itself 𝑛 times. And we know that one-fifth is a positive number. So multiplying 𝑛 positive numbers together is always going to give us a positive number. So this is strictly greater than zero. So we have shown that our sequence π‘Ž 𝑛 is nonnegative for 𝑛 greater than or equal to one.

The last thing we need to show before we use our limit comparison test is that the sequence 𝑏 𝑛 is positive for 𝑛 greater than or equal to one. We have our sequence 𝑏 𝑛 is equal to one divided by five to the 𝑛th power minus two to the 𝑛th power. And when 𝑛 is greater than or equal to one, we know that five to the 𝑛th power is bigger than two to the power. So subtracting two to the 𝑛th power from both sides of that inequality gives us five to the 𝑛th power minus two to the 𝑛th power is positive. Therefore, what we have shown is that our sequence 𝑏 𝑛 is equal to one divided by a positive number. And therefore, it must be greater zero. Therefore, we’ve shown all of our prerequisites to be true. So we can now use the limit comparison test.

The limit comparison test tells us that whether or not the series in our question diverges or converges is equivalent to whether the sum from one to ∞ of one divided by five to the 𝑛th power converges or diverges. We see that this is a geometric series. And we know that for geometric series, the sum from 𝑛 equals one to ∞ of π‘Ÿ to the 𝑛th power is equal to π‘Ÿ divided by one minus π‘Ÿ when the size of π‘Ÿ is less than one. Therefore, since we know that our series is equal to the sum from 𝑛 equals one to ∞ of one-fifth to the 𝑛th power, we can conclude that it is also equal to one-fifth divided by one minus one-fifth. Which, if we evaluate, we see is equal to a quarter.

Therefore, since that series converged, we can conclude, by using the limit comparison test, that the sum from 𝑛 equals one to ∞ of one divided by five to the 𝑛th power minus two to the 𝑛th power must converge.

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