Question Video: Using the Comparison Test Mathematics • Higher Education

Use the comparison test to decide whether the series βˆ‘_(𝑛 = 1)^(∞) (1 + (1/𝑛))^(𝑛) 𝑒^(βˆ’π‘›) is convergent or divergent.

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Video Transcript

Use the comparison test to decide whether the series the sum from 𝑛 equals one to ∞ of one plus one over 𝑛 raised to the 𝑛th power times 𝑒 to the power of negative 𝑛 is convergent or divergent.

The question gives us an infinite series. It wants us to determine whether this series is convergent or divergent by using the comparison test. To use the comparison test, we first need to decide whether we think our series is convergent or divergent. To decide this, let’s take a look at the terms of our series. Each term in this series is formed as the product of two numbers. It’s the product of one plus one over 𝑛 all raised to the 𝑛th power and 𝑒 to the power of negative 𝑛.

While not strictly necessary to answer this question, a useful piece of information to note is the limit as 𝑛 approaches ∞ of one plus one over 𝑛 raised to the 𝑛th power is actually equal to Euler’s constant 𝑒. Using this information, as 𝑛 approaches ∞, we expect the terms in our series to get closer and closer to 𝑒 times 𝑒 to the power of negative 𝑛. Which is, of course, 𝑒 to the power of one minus 𝑛. We know a series with this term is convergent. In fact, it’s a geometric series.

So, let’s try and show that our series is convergent by using the comparison test. Let’s start by recalling the convergent version of the comparison test. This 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 of our values of 𝑛. Then if we have the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is a convergent series and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all of our values of 𝑛. Then the comparison test tells us the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 must also be convergent.

We’ll start by setting π‘Ž 𝑛 to be the summand of the series given to us in the question. That’s one plus one over 𝑛 all raised to the 𝑛th power times 𝑒 to the power of negative 𝑛. And since our values of 𝑛 are greater than or equal to one, we can see this is the product of two positive numbers. So, π‘Ž 𝑛 is greater than or equal to zero for all values of 𝑛.

Now, to use the comparison test, we need to come up with a sequence 𝑏 𝑛, where 𝑏 𝑛 is greater than or equal to zero for values of 𝑛, the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges, and 𝑏 𝑛 is greater than or equal to π‘Ž 𝑛 for all of our values of 𝑛. Let’s start by recalling earlier we said as 𝑛 approached ∞, we expected that the terms in our series to get closer and closer to 𝑒 to the power of one minus 𝑛. We argued this by saying one plus one over 𝑛 all raised to the 𝑛th power approached 𝑒 as 𝑛 approached ∞.

In fact, it’s possible to show that this is an increasing sequence. So, one plus one over 𝑛 all raised to the 𝑛th power is less than 𝑒 for all of our values of 𝑛. But we don’t actually need to use this information to solve this question. And although, we could then use this information to solve this question by using the sequence 𝑏 𝑛 to be 𝑒 to the power of one minus 𝑛, there is a more simple method.

Our values of 𝑛 are integers from one to ∞. So, in particular, we know that one plus one over 𝑛 will be less than or equal to two. Raising both sides of this inequality to the 𝑛th power, we get one plus one over 𝑛 all raised to the 𝑛th power is less than or equal to two to the 𝑛th power. Using this, we’ve shown our sequence π‘Ž 𝑛 is less than or equal to two to the 𝑛th power times 𝑒 to the power of negative 𝑛.

And now, we can see why this is useful. First, by using our laws of exponents, we can write this as two over 𝑒 all raised to the 𝑛th power. Then, if we were to sum the terms of this series, we would get the sum from 𝑛 equals one to ∞ of two over 𝑒 all raised to the 𝑛th power. This a geometric series. And we know that all geometric series converge if the absolute value of their ratio is less than one. In our case, this is a geometric series with initial term two over 𝑒 and ratio of successive terms two over 𝑒. And if we calculate this, we see two over 𝑒 is approximately equal to 0.74, which is, of course, less than one. So, are serious converges.

So, we’ll set our sequence 𝑏 𝑛 in our comparison test to be two over 𝑒 all raised to the 𝑛th power. Doing this, we can see 𝑏 𝑛 is greater than or equal to zero for all of our values of 𝑛. We’ve shown the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is convergent. It’s a geometric series with ratio of successive terms approximately equal to 0.74.

Finally, we just need that π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all of our values of 𝑛. And we showed this is true because when 𝑛 is greater than or equal to one, one plus one over 𝑛 all raised to the 𝑛th power is less than or equal to two to the 𝑛th power. So, all of our prerequisites for the comparison test are true. Therefore, we can conclude that the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 must also be convergent. Therefore, by using the comparison test, we were able to show two different methods to see the sum from 𝑛 equals one to ∞ of one plus one over 𝑛 to the 𝑛th power times 𝑒 to the power of negative 𝑛 is convergent.

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