Video: Applying the 𝑛th-Term Test to Determine Whether a Given Series Is Divergent or Convergent and Determining When the Test Is Inconclusive

What can we conclude by applying the 𝑛th term divergence test in the series βˆ‘_{𝑛 = 1} ^{∞} 3𝑛/(√(6𝑛² + 4𝑛 + 5))?

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Video Transcript

What can we conclude by applying the 𝑛th term divergence test in the series the sum of three 𝑛 over the square root of six 𝑛 squared plus four 𝑛 plus five for values of 𝑛 from one to infinity?

Remember, the 𝑛th term test for divergence says that if the limit as 𝑛 approaches infinity of π‘Žπ‘› does not exist or the limit is not equal to zero, the series the sum of π‘Žπ‘› from 𝑛 equals one to infinity is divergent. And we recall that if the limit is equal to zero, we can’t tell whether the series converges or diverges and we say that the test fails.

In this question, we’re going to let π‘Žπ‘› be equal to three 𝑛 over the square root of six 𝑛 squared plus four 𝑛 plus five. And so we need to evaluate the limit of this expression as 𝑛 approaches infinity. We can’t use direct substitution. If we were to do so, we would get infinity over infinity, which we know is indeterminate. So instead, we need to find a way to manipulate our expression and see if that will help us to evaluate the limit.

To make our life just a little bit easier, let’s apply the constant multiple rule. And this is that the limit as 𝑛 approaches infinity of some constant times the function in 𝑛 is equal to that constant times the limit of the function in 𝑛. So essentially, we can take the constant three outside of our limit.

And then this next step is going to appear a little bit strange. We’re going to divide both the numerator and denominator of our expression by 𝑛. The numerator then becomes one. And the denominator becomes one over 𝑛 times the square root of six 𝑛 squared plus four 𝑛 plus five. We’re now going to take this one over 𝑛 inside our square root. To do that, we’ll need to square it. So the denominator becomes the square root of six 𝑛 squared over 𝑛 squared plus four 𝑛 over 𝑛 squared plus five over 𝑛 squared. And then, actually, this simplifies quite nicely.

We now have three times the limit as 𝑛 approaches infinity of one over the square root of six plus four over 𝑛 plus five over 𝑛 squared. And we’re now ready to apply direct substitution. As 𝑛 gets larger, four over 𝑛 and five over 𝑛 squared get smaller. They approach zero. One and six are both independent of 𝑛. So our limit becomes one over the square root of six.

We want to rationalise the denominator here. So we multiply both the numerator and denominator of our fraction by the square root of six. And we see that the limit as 𝑛 approaches infinity of π‘Žπ‘› is three times the square root of six over six, which is root six over two. We notice that this is not equal to zero. And this leads us to conclude that the series the sum of three 𝑛 over the square root of six 𝑛 squared plus four 𝑛 plus five between 𝑛 equals one and infinity is divergent.

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