Question Video: Determine the Type of Convergence or Divergence of a Series | Nagwa Question Video: Determine the Type of Convergence or Divergence of a Series | Nagwa

# Question Video: Determine the Type of Convergence or Divergence of a Series Mathematics • Higher Education

Determine if the series β_(π = 1) ^(β) (β1^(π + 1))((3^(π))/(2^(π) + 4^(π))) is absolutely convergent, conditionally convergent or divergent.

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### Video Transcript

Determine if the series the sum from π equals one to β of negative one to the power of π plus one times three to the πth power all divided by two to the πth power plus four to the πth power is absolutely convergent, conditionally convergent, or divergent.

Weβre given an infinite series, and weβre asked to determine if this series is absolutely convergent, conditionally convergent, or divergent. And the first thing thatβs always worth checking is, are the terms of our series approaching zero? And this is just another way of asking the question, βDoes our series pass the πth term divergence test?β So letβs quickly check the terms in our series are approaching zero. First, negative one to the power of π plus one is only going to change the sign of our terms. Itβs not going to change their distance from zero. Next, we can see our numerator is three to the πth power. And we know this is less than two to the πth power plus four to the πth power because in particular itβs smaller than four to the πth power.

And we can see in our denominator two to the πth power is dominated by four to the πth power. So if we ignore this, we get three to the πth power divided by four to the πth power. And that gives us a geometric sequence with ratio of successive terms three-quarters. We know this converges to zero. So our series does pass the πth term divergence test, and this gives us a method of trying to show our series is convergent. We want to compare our series to a geometric series. So weβre going to start by trying to show our series is absolutely convergent. And weβre going to do this by using the comparison test.

So weβll start by recalling the comparison test. In fact, weβre only going to recall the convergent part of the comparison test because weβre comparing our series to a geometric series with ratio of successive terms three-quarters. And we already know this is a convergent series. So the comparison test tells us if we have two series the sum from π equals one to β of π π and the sum from π equals one to β of π π, where π π and π π are both greater than or equal to zero for all of our values of π, and if we know one of our series is convergent β letβs say the sum from π equals one to β of π π is convergent β and we also have that π π is less than or equal to π π for all of our values of π, then the comparison test tells us that our sum from π equals one to β of π π must also be convergent.

And this is very similar to what we did to the series given to us in the question. Weβre going to start by checking if our series is absolutely convergent. And remember, this means we want to check for convergence or divergence of the sum from π equals one to β of the absolute value of our summand. And the reason weβre checking this first is because if we can show our series is absolutely convergent, then it must also be conditionally convergent. And we can simplify this expression. First, three to the πth power divided by two to the πth power plus four to the πth power is positive. So its absolute value is just equal to itself. And of course, negative one to the power of π plus one is always equal to negative one or one. So its absolute value will always just be equal to one.

Therefore, to check the absolute convergence of our series, we just need to check the convergence or divergence of the sum from π equals one to β of three to the πth power divided by two to the πth power plus four to the πth power. And weβll do this by using the comparison test. Weβll start by setting the summand of this series equal to π π. We already know this is greater than or equal to zero for all of our values of π. This is because each term in this expression is positive for all values of π. So their summand quotient will be positive for all values of π. Next, we want to find our sequence π π. And in fact, we already discussed how to do this when we were using the πth term divergence test.

In our denominator, four to the πth power dominates two to the πth power. And another way of putting this is if we remove two to the πth power from our denominator, weβre dividing by a smaller positive number. So we make our numbers bigger. In other words, this is less than or equal to three to the πth power divided by four to the πth power. And this will be our sequence π π. We can see from this definition this is also greater than or equal to zero for all values of π. And we chose π π and π π such that π π is less than or equal to π π for all of our values of π. To use the comparison test, all we need now is the sum from π equals one to β of π π to be convergent.

And the sum from π equals one to β of π π is equal to the sum from π equals one to β of three to the πth power divided by four to the πth power. And by using our laws of exponents, we can rewrite this as three over four all raised to the πth power. But then this is just an infinite geometric series with the ratio of successive terms π equal to three over four. And because this value of π is less than one, we know this is convergent. Therefore, by the comparison test, we must have the sum from π equals one to β of π π is also convergent. And as we showed earlier, checking whether the sum from π equals one to β of π π is convergent or divergent is the same as checking whether our series is absolutely convergent.

Therefore, by using the comparison test, we were able to show the sum from π equals one to β of negative one to the power of π plus one times three to the πth power divided by two to the πth power plus four to the πth power is absolutely convergent.

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