Video: Use the Limit Comparison Test to Determine the Convergence or Divergence of a Series

For the series βˆ‘_(𝑛 = 1)^(∞) 1/√(𝑛³) + 1, use the limit comparison test to determine whether the series converges or diverges.

04:36

Video Transcript

For the series the sum from 𝑛 equals one to ∞ of one divided by the square root of 𝑛 cubed plus one, use the limit comparison test to determine whether the series converges or diverges.

We’re given a series and we’re told we need to use the limit comparison test on this series to determine whether our series is convergent or divergent. Let’s start by recalling what the limit comparison test tells us. The limit comparison test says if we have two series the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, where our sequences π‘Ž 𝑛 and 𝑏 𝑛 are both positive for all values of 𝑛. If the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 divided by 𝑏 𝑛 is some finite positive constant 𝑐, then either both of our series are convergent or both of our series are divergent.

To start, since the question wants us to determine the convergence or divergence of the sum from 𝑛 equals one to ∞ of one divided by the square root of 𝑛 cubed plus one. We want to take a look at the terms of this series to decide if we think it will be convergent or divergent. By looking at our summand, the only part of this which changes as the value of 𝑛 changes is in our denominator. It’s the square root of 𝑛 cubed. So as our values of 𝑛 are getting larger and larger, adding one to our denominator isn’t making very much of an impact. We’d expect the terms to be very similar to one over the square root of 𝑛 cubed.

And this is a very interesting observation. We know one over the square root of 𝑛 cubed is equal to one over 𝑛 to the power of three over two. And if we were to sum terms of this form, it would be a 𝑝-series where 𝑝 is equal to three over two. This is enough motivation to try and show that our series is convergent. So let’s try using the limit comparison test. We’ll set our sequence π‘Ž 𝑛 to be one over the square root of 𝑛 cubed and our sequence 𝑏 𝑛 to be one over the square root of 𝑛 cubed plus one. It’s worth pointing out the reason we chose π‘Ž 𝑛 and 𝑏 𝑛 in this order is because our denominator of π‘Ž 𝑛 will be the most simple.

So let’s start checking the parts of the limit comparison test. First, we need π‘Ž 𝑛 to be greater than zero for all values of 𝑛 and 𝑏 𝑛 to be greater than zero for all values of 𝑛. In our case, 𝑛 is an integer and 𝑛 is greater than or equal to one. So we can see π‘Ž 𝑛 is the quotient of two positive numbers, and 𝑏 𝑛 is the quotient of two positive numbers. Therefore, both of these are positive for all values of 𝑛. Next, we need to check the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 divided by 𝑏 𝑛 is equal to some finite positive value 𝑐. And before we start evaluating this limit, remember, π‘Ž 𝑛 over 𝑏 𝑛 is actually equal to π‘Ž 𝑛 times the reciprocal of 𝑏 𝑛. So we’ll, instead, calculate the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 times the reciprocal of 𝑏 𝑛.

Substituting in our expressions for π‘Ž 𝑛 and 𝑏 𝑛 and taking the reciprocal of 𝑏 𝑛, we get the limit as 𝑛 approaches ∞ of one over the square root of 𝑛 cubed times the square root of 𝑛 cubed plus one. Next, we’ll distribute one over the square root of 𝑛 cubed over our parentheses. This gives us the limit as 𝑛 approaches ∞ of the square root of 𝑛 cubed plus one divided by the square root of 𝑛 cubed. Now, we want to divide both terms in our numerator by the square root of 𝑛 cubed. Doing this, we get the limit as 𝑛 approaches ∞ of one plus one over the square root of 𝑛 cubed.

We can now evaluate this limit directly. First, one is a constant. It doesn’t change as the value of 𝑛 changes. Next, as 𝑛 is approaching ∞, the numerator in our second term remains constant. However, the square root of 𝑛 cubed is growing without bound. Therefore, this whole term is going towards zero. Therefore, since the limit of our sum is equal to the sum of our limits, we can evaluate this limit to just give us one. This means we’ve shown the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 divided by 𝑏 𝑛 is equal to one. It’s some finite positive constant. So we’ve shown all of our prerequisites for the limit comparison test are true. Therefore, we can conclude either both of our series are convergent or both of our series are divergent.

However, remember, we chose π‘Ž 𝑛 so that it was a sequence whose terms made a 𝑝-series where 𝑝 is equal to three over two. In other words, we know that the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 is convergent. It’s a 𝑝-series where 𝑝 is equal to three over two. Therefore, by using the limit comparison test, since either both of our series are convergent or divergent, we must have both series are convergent. Therefore, we were able to show by using the limit comparison test that the sum from 𝑛 equals one to ∞ of one divided by the square root of 𝑛 cubed plus one is convergent.

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