# Video: Using the Ratio Test for Series Convergence

Using the ratio test, determine whether the series ∑_(𝑛 = 1) ^(∞) ((−2)^(𝑛) 𝑛⁴)/𝑛! is convergent, divergent, or whether the test is inconclusive.

03:42

### Video Transcript

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

Let’s begin by recalling what we mean by the ratio test. Given a series defined as the sum of 𝑎 𝑛, let’s let 𝐿 be equal to the limit as 𝑛 approaches ∞ of the absolute value of 𝑎 sub 𝑛 plus one over 𝑎 sub 𝑛. If 𝐿, the constant, is less than one, then that tells us that the series is absolutely convergent and hence convergent. If 𝐿 is greater than one, the series diverges. And if 𝐿 is equal to one, the test is inconclusive. In this case, we’re going to define 𝑎 sub 𝑛 to be equal to negative two to the 𝑛th power times 𝑛 to the fourth power all over 𝑛 factorial.

To generate 𝑎 sub 𝑛 plus one, we replace 𝑛 with 𝑛 plus one. So we get negative two to the power of 𝑛 plus one times 𝑛 plus one to the fourth power over 𝑛 plus one factorial. 𝐿 is the limit as 𝑛 approaches ∞ of the absolute value of the quotient of these. So that’s the absolute value of negative two to the power of 𝑛 plus one times 𝑛 plus one to the fourth power over 𝑛 plus one factorial divided by negative two to the 𝑛th power times 𝑛 to the fourth power over 𝑛 factorial.

Next, we recall that to divide by a fraction, we multiply by the reciprocal of that fraction. So we had the limit as 𝑛 approaches ∞ of the absolute value of negative two to the power of 𝑛 plus one times 𝑛 plus one to the fourth power over 𝑛 plus one factorial times 𝑛 factorial over negative two to the 𝑛th power times 𝑛 to the fourth power. We then recall that to divide two numbers whose bases are equal, we simply subtract the exponents. So negative two to the power of 𝑛 plus one divided by negative two to the 𝑛th power is negative two to the power of 𝑛 plus one minus 𝑛. Well, that’s negative two to the power of one or simply negative two. So we can cross cancel this negative two to the power of 𝑛.

Similarly, we know 𝑛 plus one factorial is equal to 𝑛 plus one times 𝑛 times 𝑛 minus one and so on. Or it’s equal to 𝑛 plus one times 𝑛 factorial. And we can see that we can divide through by a factor of 𝑛 factorial. And then the bit inside the limit simplifies quite nicely. It’s the absolute value of negative two times 𝑛 plus one to the fourth power over 𝑛 plus one times 𝑛 to the fourth power. We can see we can divide through by a factor of 𝑛 plus one. And we can also take out this constant factor of negative two. Remembering, of course, it needs to be the absolute value of negative two, which is simply two. So we have two times the limit as 𝑛 approaches ∞ of 𝑛 plus one cubed over 𝑛 to the fourth power.

Now, a nice way to see what happens on the next step is to distribute the parentheses. 𝑛 plus one cubed is 𝑛 cubed plus three 𝑛 squared plus three 𝑛 plus one. And we could achieve this simply by distributing the parentheses in turn or using the binomial theorem. And now, we look to manipulate this fraction somewhat. We can’t just substitute in 𝑛 equals ∞. If we do, we get ∞ over ∞, which is, of course, indeterminant. So, instead, we divide the numerator and the denominator of our fraction by the highest power of 𝑛 in the denominator, by 𝑛 to the fourth power. When we do, the denominator becomes one. And our numerator becomes 𝑛 cubed over 𝑛 to the fourth power, which is one over 𝑛, plus three 𝑛 squared over 𝑛 to the fourth power, which is three over 𝑛 squared. Plus three 𝑛 over 𝑛 to the fourth power, which is three over 𝑛 cubed, plus one over 𝑛 to the fourth power.

Now, as 𝑥 grows larger, one over 𝑛 grows smaller, three over 𝑛 squared grows smaller, three over 𝑛 cubed grows smaller, and one over 𝑛 to the fourth power grows smaller. In other words, as 𝑛 approaches ∞, each of these individual fractions approaches zero. And so, we have two times the absolute value of zero, which is simply zero. And we’ve found the value of 𝐿. Now, of course, zero is less than one. So we’re going to use the first part of the ratio test. And we say that the series converges.