Video: Using the Root Test to Determine Convergence

Consider the series βˆ‘_(𝑛 = 0)^(∞) π‘Ž_𝑛, where π‘Ž_𝑛 = ((𝑛³ + 𝑛² βˆ’ 𝑛 + 3)/(3𝑛³ + 6𝑛² + 1))^(2𝑛). Calculate lim_(𝑛 β†’ ∞) |π‘Ž_𝑛|^(1/𝑛). Hence, determine whether the series converges or diverges.

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Video Transcript

Consider the series the sum from 𝑛 equals zero to ∞ of π‘Ž sub 𝑛. Where π‘Ž sub 𝑛 equals 𝑛 cubed plus 𝑛 squared minus 𝑛 plus three all over three 𝑛 cubed plus six 𝑛 squared plus one to the power of two 𝑛. Calculate the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛 to the power of one over 𝑛. And hence, determine whether the series converges or diverges.

We’ve been told what π‘Ž sub 𝑛 is. It’s 𝑛 cubed plus 𝑛 squared minus 𝑛 plus three over three 𝑛 cubed plus six 𝑛 squared plus one, all to the power of two 𝑛. And so we need to find the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 cubed plus 𝑛 squared minus 𝑛 plus three over three 𝑛 cubed. Plus six 𝑛 squared plus one to the power of two 𝑛 all to power of one over 𝑛.

Now one of the properties is that we can take out the power of two 𝑛. So we have the limit as 𝑛 approaches ∞ of the absolute value of that fraction. Then that’s to the power of two 𝑛, which is all to the power of one over 𝑛. And that’s really useful because we can now multiply the exponents. And two 𝑛 times one over 𝑛 is just two. So we need to compute the limit as 𝑛 approaches ∞ of 𝑛 cubed plus 𝑛 squared minus 𝑛 plus three over three 𝑛 cubed plus six 𝑛 squared plus one squared.

Now we can’t do anything with direct substitution here. But to find this limit, what we can do is divide both the numerator and the denominator by the highest power of 𝑛 in our denominator. So in this case, by 𝑛 cubed. On the numerator, we then have 𝑛 cubed divided by 𝑛 cubed, which is one. The next term is 𝑛 squared over 𝑛 cubed, which simplifies to one over 𝑛. Then we have negative 𝑛 over 𝑛 cubed, which is negative one over 𝑛 squared. And then the fourth term in our numerator is three over 𝑛 cubed.

Then on the denominator, we have three 𝑛 cubed over 𝑛 cubed, which is simply three. We have six 𝑛 squared over 𝑛 cubed, which is six over 𝑛. And we have one over 𝑛 cubed.

We’re now ready to evaluate the limit as 𝑛 approaches ∞. As 𝑛 gets larger, one over 𝑛, one over 𝑛 squared, three over 𝑛 cubed, six over 𝑛, and one over 𝑛 cubed all approach zero. And so all we’re left with is one over three all squared.

Notice at this point we don’t need the absolute value signs cause we know one-third is positive. To square one-third, we square both the numerator and the denominator, which is one-ninth. And we can therefore say that the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛 to the power of one over 𝑛 is one-ninth.

The second part of this question asks us to determine whether the series converges or diverges. And in fact, we’ve set up the root test. Let’s let the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛 to the power of one over 𝑛 be equal to 𝑙. The root test tells us that if this limit 𝑙 is less than one, then the series the sum of π‘Ž sub 𝑛 is absolutely convergent and, hence, convergent. If this value is greater than one, then the series is divergent. And note that if the limit is equal to one, the root test is inconclusive. We cannot deduce anything about the convergence of our series.

It’s quite clear to us that one-ninth is less than one. According to the root test then, the series the sum from 𝑛 equals zero to ∞ of π‘Ž sub 𝑛 converges.

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