Video: Determine Whether a Quotient Series Converges or Diverges

Determine whether the series βˆ‘_(𝑛 = 1) ^(∞) 𝑛𝑒^(βˆ’π‘›) converges or diverges.

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Video Transcript

Determine whether the series from 𝑛 equals one to infinity of 𝑛 multiplied by 𝑒 to the power of negative 𝑛 converges or diverges.

The first thing we can do in these types of problems is to check if the limit of the summand as 𝑛 tends to infinity is equal to zero. This is the same as using the 𝑛th term divergence test. So in particular, this means that if the limit as 𝑛 tends to infinity of our summand π‘Ž 𝑛 is not equal to zero, then we can conclude that the series diverges. So let’s check this for the series in our question, that is, the limit as 𝑛 approaches infinity of 𝑛 multiplied by 𝑒 to the power of negative 𝑛.

We can start by writing this as a quotient. This gives us the limit as 𝑛 approaches infinity of 𝑛 divided by 𝑒 to the power of 𝑛. If we then tried to calculate this directly, we see that the limit of 𝑛 as 𝑛 approaches infinity is infinity. And the limit of 𝑒 to the power of 𝑛 as 𝑛 approaches infinity is also equal to infinity. Since we have the form of infinity over infinity, this tells us that we might be able to use L’HΓ΄pital’s rule. Which tells us if we have two differentiable functions, 𝑓 and 𝑔, and the limit of the quotient 𝑓 over 𝑔 as 𝑛 approaches π‘Ž is in an indeterminant form, then we can evaluate this limit by using the limit of 𝑓 prime over 𝑔 prime as 𝑛 approaches π‘Ž.

We know the function 𝑛 and the function 𝑒 to the 𝑛th power are both differentiable. Therefore, we can use L’HΓ΄pital’s rule to evaluate this limit. This gives us that our limit is equal to the limit as 𝑛 approaches infinity of the quotient of the derivatives. So in the numerator, we get the derivative of 𝑛, which is equal to one. And in the denominator of our limit, we get the derivative of 𝑒 to the 𝑛th power, which is just equal to 𝑒 to the 𝑛th power.

At this point, we can see that one divided by 𝑒 to the 𝑛th power is the same as saying 𝑒 to the negative 𝑛th power. Finally, we know that the limit as 𝑛 approaches infinity of 𝑒 to the power of negative 𝑛 is just equal to zero. So the limit of our summand as 𝑛 approaches infinity was equal to zero, which means that our 𝑛th term divergence test was inconclusive. This means we’re going to need to try a different method to checking the divergence or convergence of the series.

Let’s try the integral test. We recall that the integral test says that if a function 𝑓 is continuous, positive, and decreasing on an interval starting at π‘˜ all the way up to infinity, then the convergence of the series from 𝑛 equals π‘˜ to infinity of the function 𝑓 of 𝑛 is equivalent to the convergence of the integral from π‘˜ to infinity of 𝑓 of π‘₯ with respect to π‘₯.

It’s important to note that the function 𝑓 does not need to be always decreasing. It just needs to be eventually decreasing. What this means is that if we can show that our function 𝑓 is continuous and positive on this interval and eventually decreasing, then instead of discussing the convergence of the series, we can discuss the convergence of the integral.

Since the series we want to decide whether it converges or diverges is the sum from 𝑛 equals one to infinity of 𝑛 multiplied by 𝑒 to the power of negative 𝑛, we must have that π‘˜ is equal to one since this is where the series starts. And we must have that our function 𝑓 is equal to 𝑛 multiplied by 𝑒 to the power of negative 𝑛. So let’s put these values into our integral test.

We now need to show that the function 𝑛 multiplied by 𝑒 to the power of negative 𝑛 is continuous, positive, and decreasing on the interval starting at one going all the way up to infinity. We start by noticing that since π‘₯ is in the set starting at one going up to infinity, then this is the same as saying that π‘₯ is greater than or equal to one. We also know that 𝑒 to the power of negative π‘₯ is always positive.

So we can conclude that π‘₯ multiplied by 𝑒 to the negative π‘₯ is multiplying two positive numbers together. So it must also be positive. Therefore, we’ve shown that our function is positive.

Next, we know that linear functions are continuous. And we know that the exponential function is continuous. We also know that the quotient of two continuous functions is continuous on its domain. So we can use this to conclude that our function π‘₯ multiplied by 𝑒 to the power of negative π‘₯, which is equal to π‘₯ divided by 𝑒 to the power of π‘₯, is continuous on its domain. And we showed when we were checking that the function was positive that it is defined when π‘₯ is greater than or equal to one.

Therefore, we’ve shown that our function is continuous on this set. To check whether the function is decreasing, let’s consider the slope of our function, which is 𝑓 prime of π‘₯. We differentiate 𝑒 multiplied by 𝑒 to the negative π‘₯ by using the product rule. We differentiate π‘₯ to get one. And then that’s multiplied by 𝑒 to the negative π‘₯. And then we add π‘₯ multiplied by the derivative of 𝑒 to the negative π‘₯, which is negative 𝑒 to the power of negative π‘₯, which we can simplify to give us 𝑒 to the negative π‘₯ minus π‘₯𝑒 to the negative π‘₯. And if we then factor out 𝑒 to the power of negative π‘₯, we get one minus π‘₯ multiplied by 𝑒 to the power of negative π‘₯.

When π‘₯ is strictly greater than one, we notice that our slope function is a negative multiplied by a positive, which means we have a negative slope. We also know that when π‘₯ is equal to one, our slope function will be equal to zero. Therefore, when π‘₯ is equal to one, we have a local maximum. Therefore, we can conclude that our function 𝑓 is decreasing when π‘₯ is greater than or equal to one.

Since we have now shown the three prerequisites for using the integral test, we can now use this integral test to determine the convergence of the series in our question as the convergence of an integral instead. So let’s try to determine the convergence of our integral, that is, the limit as 𝑑 approaches infinity of the integral from one to 𝑑 of 𝑑 multiplied by 𝑒 to the power of negative 𝑑 with respect to 𝑑.

We can evaluate this integral by using integration by parts, which says that the integral of 𝑒 multiplied by d𝑣 d𝑑 with respect to 𝑑 is equal to 𝑒 multiplied by 𝑣 minus the integral of 𝑣 multiplied by d𝑒 d𝑑 with respect to 𝑑. So let’s start by setting 𝑒 equal to 𝑑 and d𝑣 d𝑑 is equal to 𝑒 to the power of negative 𝑑. We differentiate 𝑒 with respect to 𝑑 to get that d𝑒 d𝑑 is equal to one. And if d𝑣 d𝑑 is equal to 𝑒 to the power of negative 𝑑, then the antiderivative is just negative 𝑒 to the power of negative 𝑑.

So substituting this information into our integration by parts formula gives us the limit as 𝑑 approaches infinity of negative 𝑑 multiplied by 𝑒 to the negative 𝑑 evaluated from one to 𝑑 minus the integral from one to 𝑑 of negative 𝑒 to the power of negative 𝑑 with respect to 𝑑. Evaluating negative 𝑑 multiplied by 𝑒 to the negative 𝑑 at one and 𝑑 gives us negative 𝑑 multiplied by 𝑒 to the negative 𝑑 plus 𝑒 to the power of negative one.

And we also know the antiderivative of negative 𝑒 to the power of negative 𝑑 is just 𝑒 to the power of negative 𝑑. So in our limit, we subtract 𝑒 to the power of negative 𝑑 evaluated between one and 𝑑. Evaluating 𝑒 to the power of negative 𝑑 at one and 𝑑 gives us 𝑒 to the power of negative 𝑑 minus 𝑒 to the power of negative one. Which we can simplify to give us the limit as 𝑑 approaches infinity of negative 𝑑 multiplied by 𝑒 to the power of negative 𝑑 plus two multiplied by 𝑒 to the power of negative one minus 𝑒 to the power of negative 𝑑.

We can evaluate each part of this limit separately. We saw earlier when we did the 𝑛th term divergence test on our series that the limit of 𝑛 multiplied by 𝑒 to the power of negative 𝑛 as 𝑛 approaches infinity is equal to zero. So the first term in our limit just limits to zero. We know that two multiplied by 𝑒 to the power of negative one is just a constant. So its limit is equal to itself. Also, we used earlier that the limit as 𝑛 approaches infinity of 𝑒 to the power of negative 𝑛 is equal to zero. So we know that this limit is equal to zero.

Therefore, we have shown that this integral is convergent. In fact, it’s equal to two multiplied by 𝑒 to the power of negative one. Since the integral is finite, we can conclude by using the integral test that our series must converge, giving us that the sum as 𝑛 equals one to infinity of 𝑛 multiplied by 𝑒 to the power of negative 𝑛 must converge.

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