Question Video: Using the 𝑝-Series Test | Nagwa Question Video: Using the 𝑝-Series Test | Nagwa

# Question Video: Using the π-Series Test Mathematics • Higher Education

Use the π-series test to determine whether the series β_(π = 1)^(β) β(4πβ΄)/(5π^(5/2)) is divergent or convergent.

04:22

### Video Transcript

Use the π-series test to determine whether the series the sum from π equals one to β of the cube root of four π to the fourth power divided by five π to the power of five over two is divergent or convergent.

The question gives us an infinite series. It wants us to determine whether this series is divergent or convergent by using the π-series test. Letβs start by recalling what we mean by the π-series test.

To start, a π-series is a series of the form the sum from π equals one to β of one divided by π to the πth power. We know this is convergent for all values of π greater than one and divergent for all values of π less than or equal to one. This means if we can compare a series to a π-series, we can determine whether theyβre convergent or divergent. This is called the π-series test.

So we want to manipulate the series given to us in the question so that we can directly compare it with a π-series. To do this, weβre going to want to use our laws of exponents. Letβs start by rewriting the numerator of our summand. Weβll do this by using two of our laws of exponents. First, the cube root of a number is that number raised to the power of one-third.

Next, we can distribute an exponent over a product. π times π all raised to the power of one-third is equal to π to the power of one-third times π to the power of one-third. Doing this, we can rewrite the numerator of our summand as four to the power of one-third times π to the fourth power raised to the power of one-third. And we can see that we can simplify this even further by using our laws of exponents.

This time, weβre going to use the fact that π to the power of π all raised to the power of π is just equal to π to the power of π times π. This will let us write π to the fourth power all raised to the power of one-third as π to the power of four over three.

So by applying these laws of exponents to our series, we were able to rewrite it as the sum from π equals one to β of four to the power of one-third times π to the power of four over three divided by five π to the power of five over two. We can now see that both the numerator and the denominator of our summand contain π raised to some power.

Once again, we can simplify this by using our laws of exponents. We know π to the power of π divided by π to the power of π is equal to π to the power of π minus π. This means we can simplify π to the power of four over three divided by π to the power of five over two as π to the power of four over three minus five over two. And we can then simplify this. Four-thirds minus five over two is equal to negative seven divided by six.

So this means weβve now written our series as the sum from π equals one to β of four to the power of one-third times π to the power of negative seven over six divided by five. We can see this is looking more and more like a π-series. We can see in the definition of a π-series the term of π is in the denominator.

However, in our series, we have this in the numerator. So weβll want to move this into the denominator. And to do this, weβll once again use our laws of exponents. We know π to the power of negative π is just equal to one over π to the power of π. So by using this, weβve now rewritten our series as the sum from π equals one to β of four to the power of one-third divided by five π to the power of seven over six.

We can now see in our general π-series that the numerator is just equal to one. However, in our series, our numerator is four to the power of one-third. And in our denominator, we have a coefficient of five. But these are both constant factors. So we can take these outside of our series.

So by taking out our constant factor of four to the power of one-third divided by five from our series. We get four to the power of one-third over five times the sum from π equals one to β of one divided by π to the power of seven over six. And this series is a π-series. In fact, itβs a π-series where π is equal to seven over six, which we know is greater than one. So this means that this π-series is convergent.

So letβs think about what weβve just shown. Weβve shown the series given to us in the question is a constant multiple of a convergent series. That means the series given to us in the question must also be convergent.

Therefore, by using the π-series test, we were able to show the sum from π equals one to β of the cube root of four π to the fourth power divided by five π to the power of five over two is a convergent series.

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