Video: Determine Whether a P-Series Is Convergent or Divergent

Determine whether the series βˆ‘ _(𝑛 = 1) ^(∞) 1/(π‘›βˆš(𝑛³)) converges or diverges.

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Video Transcript

Determine whether the series the sum from 𝑛 equals one to ∞ of one divided by 𝑛 times the square root of 𝑛 cubed converges or diverges.

We’re given a series, and we’re asked to determine whether this series converges or diverges. And there’s a lot of different ways of dealing with this. For example, we need to check that our terms are getting closer and closer to zero. This is called the 𝑛th term divergence test. We need to check the limit as 𝑛 approaches ∞ our summand approaches zero. And in this case, this is true. We can see this because our numerator remains constant; however, our denominator is growing without bound.

And at this point, there’s a few different things we could try. We could start looking at our partial sums; however, in this case, there’s actually an easier method. If we look at our summand, we can see we can rearrange this by using our laws of exponents. And in this case, we will be able to rewrite this as a 𝑝-series. So, we’ll use the 𝑝-series test to determine the convergence or divergence of this series.

So, let’s start by recalling the 𝑝-series test. This tells the sum from 𝑛 equals one to ∞ of one divided by 𝑛 to the power of 𝑝 is convergent if 𝑝 is greater than one and divergent if 𝑝 is less than or equal to one. To use this, we’re going to need to use our laws of exponents to rewrite our series. First, recall taking the square root of a number is the same as raising that number to the power of one-half. So, the square root of 𝑛 cubed is 𝑛 cubed raised to the power of one-half. But we can then simplify this even further. 𝑛 cubed all raised to the power of one-half is equal to 𝑛 to the power of three times one-half, which is 𝑛 to the power of three over two.

So, we were able to rewrite our series as the sum from 𝑛 equals one to ∞ of one divided by 𝑛 times 𝑛 to the power of three over two. But we can simplify this even further by using our laws of exponents. By writing 𝑛 as 𝑛 to the first power, we can multiply 𝑛 to the first power by 𝑛 to the power of three over two by adding our exponents. Doing this, we get the sum from 𝑛 equals one to ∞ of one divided by 𝑛 to the power of one plus three over two. And, of course, we can simplify this. One plus three over two is equal to five over two.

So, we were able to rewrite the power series given to us in the question as the sum from 𝑛 equals one to ∞ of one divided by 𝑛 to the power of five over two. And we can see this is exactly equal to a 𝑝-series, where the value of 𝑝 is equal to five over two and five over two is greater than one. So, by our 𝑝-series test, this means that our series must be convergent.

Therefore, by using our laws of exponents, we were able to show the sum from 𝑛 equals one to ∞ of one divided by 𝑛 times the square root of 𝑛 cubed is a 𝑝-series where 𝑝 is equal to five over two. And then, by using the 𝑝-series test, we can conclude that this means our series must be convergent.

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