Video: Determining Whether a Given Series Is Absolutely Convergent- Conditionally Convergent- or Divergent

Consider the series βˆ‘_(𝑛 = 0)^(∞) (βˆ’1)^(𝑛)/(2𝑛 + 1). Decide whether the series is absolutely convergent, conditionally convergent, or divergent.

06:09

Video Transcript

Consider the series the sum from 𝑛 equals zero to ∞ of negative one raised to the 𝑛th power divided by two 𝑛 plus one. Decide whether the series is absolutely convergent, conditionally convergent, or divergent.

The question gives us a series and asks us to determine the different types of convergence and divergence that this series has. One thing that we might question is which order to check these conditions. We recall that if the sum from 𝑛 equals zero to ∞ is not conditionally convergent, then, in particular, we could conclude that it must not be absolutely convergent. And this is because all absolutely convergent sequences are conditionally convergent. Using this as our basis, we will follow the following flow chart. First, we will check if the series is conditionally convergent. If we know the series is conditionally convergent, then it must not be divergent. So we only need to check for absolute convergence.

Similarly, if we show the series is not conditionally convergent, then it cannot be absolutely convergent. So we only need to check for divergence. To check whether the series given to us in the question is conditionally convergent, we recall the alternating series test. Which tells us if a sequence π‘Ž 𝑛 is equal to negative one to the 𝑛th power multiplied by 𝑏 𝑛, where our sequence 𝑏 𝑛 is eventually positive and eventually decreasing. Then, we can conclude that the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 converges.

We see that the series given to us is an alternating series. Its summand can be rewritten as negative one to the 𝑛th power multiplied by one divided by two 𝑛 plus one. So, we set our sequence 𝑏 𝑛 to be equal to one divided by two 𝑛 plus one in our alternating series test. And we show that 𝑏 𝑛 is eventually positive and eventually decreasing. Then, we would be able to conclude using the alternating series test that the series given to us in our question is conditionally convergent.

The first thing we notice is that our sequence 𝑏 𝑛 must be positive since if 𝑛 is bigger than zero, then two multiplied by 𝑛 is a positive number, we add one, that’s also a positive number, and then the reciprocal of this must also be a positive number. The next thing we notice is that as we’re increasing the value of 𝑛, we’re actually increasing the value of our denominator. So, our sequence 𝑏 𝑛 is getting smaller.

Therefore, we can conclude by using our alternating series test the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power divided by two 𝑛 plus one converges. Since we have an alternating series given to us in the question, what we have shown is that our series is conditionally convergent. Since we have shown that our series is conditionally convergent, it cannot be divergent. Therefore, the only thing left to check is absolute convergence.

And we recall that checking the absolute convergence is the same as checking whether the sum from 𝑛 equals zero to ∞ of the absolute value of our sequence π‘Ž 𝑛 converges. Let’s see what the absolute value of our summand is. That is, what is the absolute value of negative one to the 𝑛th power divided by two 𝑛 plus one? The first thing we do is split the absolute value over our quotient. So, we have the absolute value of negative one to the 𝑛th power divided by the absolute value of two 𝑛 plus one. We see that negative one raised to the 𝑛th power will always output a negative one or one. So, just taking the absolute value of these outputs is always one. We know that two 𝑛 plus one is always positive since 𝑛 is greater than or equal to zero. Therefore, we have shown that the absolute value of our summand is equal to one divided by two 𝑛 plus one.

So, what we have shown is that, to check the absolute convergence of our series, we need to check the convergence of the series, which is the sum from 𝑛 equals zero to ∞ of one divided by two 𝑛 plus one. Our summand is just one divided by a linear factor. So, it makes sense to compare this to the harmonic series, which we will shift over by one to make the limits of our sum match. In fact, we’re going to use the limit comparison test. We recall that the limit comparison test tells us that if two sequences π‘Ž 𝑛 and 𝑏 𝑛 are eventually positive and if the limit as 𝑛 approaches ∞ of the quotient π‘Ž 𝑛 over 𝑏 𝑛 is positive. Then, we can conclude that the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 and the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 either both converge or they both diverge.

The first thing we notice is that both of our sequences, one divided by 𝑛 plus one and one divided by two 𝑛 plus one, are positive when 𝑛 is greater than or equal to zero. So, we’ve shown the first part of our limit comparison test is true. Now, all we need to do is check the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 divided by 𝑏 𝑛. Substituting in π‘Ž 𝑛 is equal to one divided by 𝑛 plus one and 𝑏 𝑛 is equal to one divided by two 𝑛 plus one gives us the limit as 𝑛 approaches ∞ of one divided by 𝑛 plus one all divided by one divided by two 𝑛 plus one. Instead of dividing by the fraction one divided by two 𝑛 plus one, we can instead multiply by the reciprocal. This gives us the limit as 𝑛 approaches ∞ of two 𝑛 plus one all divided by 𝑛 plus one.

One way of evaluating limits like this is to divide both the numerator and the denominator by 𝑛. Dividing the numerator of the fraction in our limit gives us two plus one divided by 𝑛. And dividing the denominator of the function in our limit by 𝑛 gives us one plus one divided by 𝑛. Since our limit is as 𝑛 approaches ∞, the reciprocal function one divided by 𝑛 is approaching zero. Therefore, we can evaluate this limit as the limit as 𝑛 approaches ∞ of two divided by one. Since this is just a constant, our limit is equal to two. Therefore, we’ve shown the second part of our limit comparison test is true. So, by using the conclusion of the limit comparison test, we have the sum from 𝑛 equals zero to ∞ of one divided by two 𝑛 plus one and the sum from 𝑛 equals zero to ∞ of one divided by 𝑛 plus one either both converge or they both diverge.

However, we chose our second series to just be the harmonic series, the sum from 𝑛 equals one to ∞ of one divided by 𝑛. And in particular, we know the infinite sum of the harmonic series diverges. So, by the limit comparison test, we’ve shown that the sum from 𝑛 equals zero to ∞ of one divided by two 𝑛 plus one must diverge. Since this series is equal to the absolute value of the series given to us in the question, we can conclude that the series given to us in the question is not absolutely convergent. Therefore, what we have shown is the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power all divided by two 𝑛 plus one is conditionally convergent.

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