Video: Using the Ratio Test to Determine the Convergence of a Series

Using the ratio test, determine whether the series βˆ‘_(𝑛 = 1)^(∞) 2𝑛^(𝑛)/𝑛! is convergent, divergent, or whether the test is inconclusive.

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Video Transcript

Using the ratio test, determine whether the series the sum from 𝑛 equals one to ∞ of two times 𝑛 to the 𝑛th power divided by 𝑛 factorial is convergent, divergent, or whether the test is inconclusive.

We’re given a series, and we need to use the ratio test to determine whether this series is convergent, divergent, or whether the ratio test is inconclusive. So let’s start by recalling the ratio test. The ratio test tells us, for a series the sum from 𝑛 equals one to ∞ of π‘Žπ‘›, where we set 𝑙 equal to the limit as 𝑛 approaches ∞ of the absolute value of π‘Žπ‘› plus one divided by π‘Žπ‘›, then if this value of 𝑙 is less than one, then the ratio test tells us that our series must be absolutely convergent. However, if this value of 𝑙 is greater than one, then the ratio test tells us that our series must be divergent. Finally, if this value of 𝑙 is equal to one, then the ratio test is inconclusive.

So to apply the ratio test, we need to calculate this value of 𝑙. To do this, we need to calculate the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms in the series given to us in the question. So we’ll set our sequence π‘Žπ‘› to be the summand of our series. That’s two times 𝑛 to the 𝑛th power divided by 𝑛 factorial. And we need to calculate 𝑙, which is the limit as 𝑛 approaches ∞ of the absolute value of π‘Žπ‘› plus one divided by π‘Žπ‘›. However, before we evaluate this limit, we’ll make one small adjustment. Instead of dividing by π‘Žπ‘›, we’ll multiply by the reciprocal of π‘Žπ‘›.

We’re now ready to start calculating the value of 𝑙. We need to substitute 𝑛 plus one and 𝑛 to find our values of π‘Žπ‘› plus one and π‘Žπ‘›. This gives us the limit as 𝑛 approaches ∞ of the absolute value of two times 𝑛 plus one raised to the power of 𝑛 plus one divided by 𝑛 plus one factorial times 𝑛 factorial divided by two times 𝑛 to the 𝑛th power, where it’s worth pointing out we evaluated the reciprocal of π‘Žπ‘› by inverting our fraction.

We’re now ready to start simplifying. First, in our denominator, we can rewrite 𝑛 plus one factorial as 𝑛 plus one multiplied by 𝑛 factorial. This then means we can cancel the shared factor of 𝑛 factorial in our numerator and our denominator. But notice we also have 𝑛 plus one factors of 𝑛 plus one in our numerator. So we can also cancel one of our factors of 𝑛 plus one. Finally, we can also cancel the shared factor of two in our numerator and our denominator. This gives us the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one all raised to the 𝑛th power divided by 𝑛 raised to the 𝑛th power.

To simplify this expression, we’re going to need to recall one of our laws of exponents. π‘Ž to the power of 𝑏 divided by 𝑐 to the power of 𝑏 is actually equal to π‘Ž over 𝑐 all raised to the power of 𝑏. And we can see we’re dividing two terms both raised to the 𝑛th power. This gives us the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one all divided by 𝑛 all raised to the 𝑛th power.

Now, all we need to do to evaluate this limit is to evaluate the division by 𝑛. We’ll just divide both terms in our numerator by 𝑛. This gives us the limit as 𝑛 approaches ∞ of the absolute value of one plus one over 𝑛 all raised to the 𝑛th power. And this should now start reminding us of a limit result we know.

We need to remember the limit as 𝑛 approaches ∞ of one plus one over 𝑛 all raised to the 𝑛th power is equal to Euler’s constant 𝑒. This is almost exactly what we have. We’re just taking the absolute value of this limit. But remember, our values of 𝑛 range from one and then increase. So one plus one over 𝑛 is positive. And if one plus one over 𝑛 is positive for all of our values of 𝑛, then we also know one plus one over 𝑛 all raised to the 𝑛th power is positive for all of our values of 𝑛. So we can remove the absolute value signs. So this limit evaluates to give us 𝑒.

We found that our value of 𝑙 is equal to 𝑒. And we know 𝑒 is approximately equal to 2.7. This means our value of 𝑙 is one. And that means the ratio test is conclusive. In fact, the ratio test tells us that our series must be divergent.

Therefore, by applying the ratio test to the series the sum from 𝑛 equals one to ∞ of two times 𝑛 to the 𝑛th power divided by 𝑛 factorial, we were able to show that this series is divergent.

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