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

Using the 𝑛th-term test, determine whether the series βˆ‘_(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 π‘Ž 𝑛 is not equal to zero, then the infinite sum of π‘Ž 𝑛 diverges. And if the limit as 𝑛 approaches ∞ of the sequence π‘Ž 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 π‘Ž 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 π‘Ž 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 π‘₯ to the 𝑛th power divided by 𝑦 to the 𝑛th power is equal to π‘₯ 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 π‘Ÿ is less than one, then the limit as 𝑛 approaches ∞ of π‘Ÿ to the 𝑛th power is equal to zero. And if the absolute value of our ratio π‘Ÿ is greater than one, then the limit as 𝑛 approaches ∞ of π‘Ÿ to the 𝑛th power does not exist. In our limit, we have that π‘Ÿ 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 π‘Ÿ is greater than one, then the limit as 𝑛 approaches ∞ of π‘Ÿ 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 π‘Ž 𝑛 as 𝑛 approaches ∞. So in particular, we’ve shown that the limit of π‘Ž 𝑛 as 𝑛 approaches ∞ is not equal to zero. Therefore, using our 𝑛th-term divergence test, we must have that the infinite sum of π‘Ž 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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.