# Video: Transcendental Functions as Power Series

Consider the series ∑_(𝑛 = 1)^(∞) (𝑛³/(3𝑛!)). Determine whether the series converges or diverges.

02:57

### Video Transcript

Consider the series the sum from 𝑛 is equal to one to ∞ of 𝑛 cubed over three 𝑛 factorial. Determine whether the series converges or diverges.

We can try using the ratio test to determine whether the series converges or diverges. The ratio test tells us that for some series, which is the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 with, 𝐿 being equal to the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛, that firstly, if 𝐿 is less than one, then our series converges absolutely. Secondly, if 𝐿 is greater than one, then our series diverges. And finally, if 𝐿 is equal to one, then the test is inconclusive. Now, in our case, we can see that 𝑎 𝑛 is equal to 𝑛 cubed over three 𝑛 factorial. Let’s quickly note that this three 𝑛 factorial in the denominator is equal to three multiplied by 𝑛 factorial. And so, the factorial symbol affects only the 𝑛 and not the three.

Next, we can write down what 𝑎 𝑛 plus one is. We simply change each 𝑛 in 𝑎 𝑛 to 𝑛 plus one. And we see that 𝑎 𝑛 plus one is equal to 𝑛 plus one cubed over three 𝑛 plus one factorial. Now, we’re ready to find 𝐿. We have that 𝐿 is equal to the limit as 𝑛 tends to ∞ of the absolute value of 𝑎 𝑛 plus one over 𝑎 𝑛. We substitute in 𝑎 𝑛 plus one and one over 𝑎 𝑛. Now, we’re ready to start simplifying.

First, we can cross cancel the factor of three. Next, we know that we can rewrite 𝑛 plus one factorial as 𝑛 plus one multiplied by 𝑛 factorial. This will allow us to write our limit as the limit as 𝑛 tends to ∞ of the absolute value of 𝑛 plus one cubed multiplied by 𝑛 factorial over 𝑛 plus one multiplied by 𝑛 factorial multiplied by 𝑛 cubed. And so, we can cross out the factor of 𝑛 factorial from the numerator and denominator. We can also cancel a factor of 𝑛 plus one from the denominator with one of the factors of 𝑛 plus one from the numerator. What we’re left with here is the limit as 𝑛 tends to ∞ of the absolute value of 𝑛 plus one squared over 𝑛 cubed.

Next, we can distribute the square in the numerator. Here, it is clear that our highest power of 𝑛 in the numerator is 𝑛 squared and our highest power of 𝑛 in the denominator is 𝑛 cubed. Since we’re taking the limit as 𝑛 tends to ∞ and the highest power of 𝑛 in the denominator is greater than the highest power of 𝑛 in the numerator, this tells us that this limit will be equal to zero. What we have found here is that our value of 𝐿 for this series is zero and zero is less than one. So when we look at our rule for the ratio test, we see that we’ve satisfied condition number one. And this tells us our solution to the question, which is that the sum from 𝑛 equals one to ∞ of 𝑛 cubed over three 𝑛 factorial converges absolutely.