# Video: Determining If a Given Series Is Divergent Using the 𝑛th Term Test

Using the 𝑛th-term test, determine whether the series ∑_(𝑛 = 0)^(∞) 4^(𝑛)/3^(𝑛) is divergent or the test fails.

02:35

### Video Transcript

Using the 𝑛th-term test, determine whether the series the sum from 𝑛 equals zero to ∞ of four to the 𝑛th power divided by three to the 𝑛th power is divergent or the test fails.

The question wants us to use the 𝑛th-term test on the series the sum from 𝑛 equals zero to ∞ of four to the 𝑛th power divided by three to the 𝑛th power. And we recall that the 𝑛th-term divergence test tells us that if the limit as 𝑛 approaches ∞ of the sequence 𝑎 𝑛 is not equal to zero, then the infinite sum of 𝑎 𝑛 diverges. And if the limit as 𝑛 approaches ∞ of the sequence 𝑎 𝑛 is equal to zero, then the test is inconclusive.

Since the question wants us to use this test on the series the sum from 𝑛 equals zero to ∞ of four to the 𝑛th power divided by three to the 𝑛th power, we’ll set 𝑎 𝑛 to be equal to four to the 𝑛th power divided by three to the 𝑛th power. So let’s now calculate the limit as 𝑛 approaches ∞ of our sequence 𝑎 𝑛. That’s equal to the limit as 𝑛 approaches ∞ of four to the 𝑛th power divided by three to the 𝑛th power.

Next, we’re going to use that 𝑥 to the 𝑛th power divided by 𝑦 to the 𝑛th power is equal to 𝑥 divided by 𝑦 all raised to the 𝑛th power to rewrite our limit. This gives us the limit as 𝑛 approaches ∞ of four-thirds all raised to the 𝑛th power. This gives us the limit as 𝑛 approaches ∞ of some constant raised to the 𝑛th power. This is a geometric sequence.

And we recall that if the absolute value of our ratio 𝑟 is less than one, then the limit as 𝑛 approaches ∞ of 𝑟 to the 𝑛th power is equal to zero. And if the absolute value of our ratio 𝑟 is greater than one, then the limit as 𝑛 approaches ∞ of 𝑟 to the 𝑛th power does not exist. In our limit, we have that 𝑟 is equal to four divided by three, which is greater than one. And using what we know about geometric sequences, we have if the absolute value of our ratio 𝑟 is greater than one, then the limit as 𝑛 approaches ∞ of 𝑟 to the 𝑛th power does not exist.

So we have that the limit as 𝑛 approaches ∞ of four-thirds to the 𝑛th power does not exist. But this limit is equal to our limit of 𝑎 𝑛 as 𝑛 approaches ∞. So in particular, we’ve shown that the limit of 𝑎 𝑛 as 𝑛 approaches ∞ is not equal to zero. Therefore, using our 𝑛th-term divergence test, we must have that the infinite sum of 𝑎 𝑛 diverges.

So in conclusion, we’ve shown if we use the 𝑛th-term divergence test on the series the sum from 𝑛 equal zero to ∞ of four to the 𝑛th power divided by three to the 𝑛th power, then we can conclude that this series is divergent.