Video: The 𝑛th Term Divergence Test

What can we conclude by applying the 𝑛th term divergence test in the series βˆ‘ _(𝑛 = 1) ^(∞) ((5𝑛 + 3)^(2𝑛))/((7𝑛² + 2)^(𝑛))?

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Video Transcript

What can we conclude by applying the 𝑛th term divergence test in the series? The sum from 𝑛 equals one to ∞ of five 𝑛 plus three to the power of two 𝑛 over seven 𝑛 squared plus two to the power of 𝑛.

We begin by recalling what we mean by the 𝑛th term divergence test. The 𝑛th term test for divergence says that if the limit as 𝑛 approaches ∞ of π‘Ž sub 𝑛 does not exist or is not equal to zero. Then the series the sum from 𝑛 equals one to ∞ of π‘Ž sub 𝑛 is divergent. We recall also 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 or it’s inconclusive.

So let’s begin by letting π‘Ž sub 𝑛 be equal to five 𝑛 plus three to the power of two 𝑛 over seven 𝑛 squared plus two to the power of 𝑛. We’re going to evaluate the limit of this as 𝑛 approaches ∞. Now we can’t apply direct substitution. If we do, we get ∞ over ∞, which we know to be undefined. So we’re going to manipulate the limit a little bit.

We’re going to begin by writing it as the limit as 𝑛 approaches ∞ of five 𝑛 plus three squared over seven 𝑛 squared plus two all to the power of 𝑛. And remember, we’re allowed to do this because when we have the brackets, we multiply the exponents. So when we multiply two by 𝑛, we get two 𝑛 as required. And then, on the denominator, we simply have seven 𝑛 squared plus two to the 𝑛th power.

We’re now going to distribute the parentheses on the numerator of our limit. Five 𝑛 plus three squared is five 𝑛 plus three times five 𝑛 plus three. Well, that gives us 25𝑛 squared plus 30𝑛 plus nine. We still can’t apply direct substitution. So what we’re going to do is we’re going to look at the numerator and denominator of the limit.

One trick we have for evaluating limits of this form is to divide both the numerator and the denominator by the highest power of 𝑛 in the denominator. So we’re going to divide everything through by 𝑛 squared. Let’s do that term by term.

On the numerator, we have 25𝑛 squared over 𝑛 squared, which is 25. We then have 30𝑛 over 𝑛 squared, which simplifies to 30 over 𝑛. And we have nine over 𝑛 squared. Then on the denominator, we have seven 𝑛 squared over 𝑛 squared plus two over 𝑛 squared, which simplifies to seven plus two over 𝑛 squared.

So we’re now looking to evaluate the limit as 𝑛 approaches ∞ of 25 plus 30 over 𝑛 plus nine over 𝑛 squared all over seven plus two over 𝑛 squared all to the 𝑛th power. Now in fact, as 𝑛 grows large, it should be quite clear that 30 over 𝑛, nine over 𝑛 squared, and two over 𝑛 squared grow ever smaller. In fact, as 𝑛 approaches ∞, each of these fractions themselves approach zero.

And so that leaves us with 25 over seven to the 𝑛th power. But of course, as 25 over seven is a fraction which is greater than one, we then find that as 𝑛 approaches ∞, 25 over seven to the 𝑛th power also approaches ∞. And so the limit as 𝑛 approaches ∞ of π‘Ž sub 𝑛 is ∞. This is not equal to zero. So by the 𝑛th term divergence test, our series diverges.

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