Video: Using the Root Test to Determine Convergence

Consider the series βˆ‘_(𝑛 = 0)^(∞) π‘Ž_𝑛, where π‘Ž_𝑛 = (𝑛 + 1)^(𝑛)/6^(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 equal zero to ∞ of π‘Ž sub 𝑛, where π‘Ž sub 𝑛 is equal to 𝑛 plus one to the 𝑛th power over six 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 have had π‘Ž sub 𝑛 defined for us. It’s 𝑛 plus one to the 𝑛th power over six to the power of two 𝑛. And so, to answer the first part of this question, we need to establish the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one to the 𝑛th power over six to the power of two 𝑛 all to the power of one over 𝑛.

Now, in fact, to make this a little bit easier, we’re going to rewrite the part inside the absolute values signs. We’re going to write it as 𝑛 plus one over six squared all to the 𝑛th power. Then, we can take that 𝑛th power outside the absolute value signs. And we have the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one over six squared to the 𝑛th power all to the power of one over 𝑛.

Now, when we have the brackets, we multiply the exponents. And 𝑛 times one over 𝑛 is simply one. And so, we’re looking to find the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one over six squared. Now, of course, one over six squared is independent of 𝑛. So, we can take that outside of our limit. And now, we can see we’re looking to compute one over six squared times the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one.

One over six squared is one over 36. But as 𝑛 gets larger, the absolute value of 𝑛 plus one gets larger. In other words, it also approaches ∞. So, we have one over 36 times ∞, which is also ∞. And so, we calculated the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛 to the power of one over 𝑛. It’s ∞.

The second part of this question asks us to determine whether the series converges or diverges. Well, what we’ve actually set up for is to use the root test. Let’s call the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛, all to the power of one over 𝑛, 𝑙. If this value of 𝑙 is less than one, then our series, the sum of π‘Ž sub 𝑛, is absolutely convergent and, therefore, convergent. If 𝑙 is greater than one, the series diverges. And actually, if it’s equal to one, the root test is inconclusive. Well, ∞ is quite clearly greater than one. And so, we deduce that the series diverges.

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