Question Video: Determine Whether a Series Is Convergent or Divergent | Nagwa Question Video: Determine Whether a Series Is Convergent or Divergent | Nagwa

# Question Video: Determine Whether a Series Is Convergent or Divergent Mathematics • Higher Education

Use the π-series test to determine whether the series β_(π = 1) ^(β) π^(β10)/π^(β10) is divergent or convergent.

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### Video Transcript

Use the π-series test to determine whether the series the sum from π equals one to β of π to the power of the cube root to 10 divided by π to the power of the square root of 10 is divergent or convergent.

The question gives us an infinite series. And it wants us to determine whether the series is convergent or divergent by using the π-series test. So the first thing weβre going to need to recall is what the π-series test tells us. First, a π-series is a series of the form the sum from π equals one to β of one divided by π to the power of π. And we know this will always be convergent if the value of π is bigger than one, and it will always be divergent if the value of π is less than or equal to one.

And the π-series test means we just need to compare our series with a π-series. In this case, weβre going to want to rewrite our series to be similar to a π-series. So letβs start with our series the sum from π equals one to β of π to the power of the cube root of 10 divided by π to the power of the square root of 10. We want to try and write this as close to our π-series as possible. We want to write the summand as one divided by π to the power π for some constant π. And we can simplify our summand by using our laws of exponents.

First, recall by using our laws of exponents, π to the power of π divided by π to the power of π will be equal to π to the power of π minus π. We can apply this to our summand by setting our value of π equal to π, π equal to the cube root of 10, and π equal to the square root of 10. This gives us that our series is equal to the sum from π equals one to β of π to the power of the cube root of 10 minus the square root of 10. And now we can see this is almost a π-series. We just need to use one more of our laws of exponents.

We recall π to the power of π is the same as one divided by π to the power of negative π. We can apply this to our summand. We set our value of π equal to π and π equal to the cube root of 10 minus the square root of 10. And doing this gives us the sum from π equals one to β of one divided by π to the power of negative one times the cube root of 10 minus the square root of 10. And now we can see weβve written this exactly in the form of a π-series, where our value of π is the exponent of π. In other words, weβve shown the series given to us in the question is a π-series, where the value of π is negative one times the cube root of 10 minus the square root of 10.

And we can calculate this value of π is equal to 1.0078 to four decimal places. The important thing to notice here is our value of π is greater than one. And we know that a π-series with a value of π greater than one must be convergent. So our series must be convergent. Therefore, because we were able to show the series given to us in the question the sum from π equals one to β of π to the power of the cube root of 10 divided by π to the power of the square root of 10 is a π-series where π is approximately equal to 1.0078, we can conclude from the π-series test that our series must be convergent.

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