Video: Find the Interval of Convergence for a Power Series

Find the interval of convergence for the power series βˆ‘_(𝑛 = 0) ^(∞) (π‘₯^(3𝑛+ 2)/(5𝑛 + 10)).

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Video Transcript

Find the interval of convergence for the power series the sum from 𝑛 equal zero to ∞ of π‘₯ to the power of three 𝑛 plus two all divided by five 𝑛 plus 10.

The question asks us to find the interval of convergence for this power series. We recall that we call 𝐼 an interval of the real numbers, the interval of convergence for a series of π‘Žπ‘›. If π‘₯ is in the interval 𝐼, then the sum of π‘Žπ‘› converges. And if π‘₯ is not in the interval 𝐼, then the sum of π‘Žπ‘› must diverge. So we’re looking for which values of π‘₯ will cause the sum from 𝑛 equals zero to ∞ of π‘₯ to the power of three 𝑛 plus two all divided by five 𝑛 plus 10 to converge.

One of the best ways to find an interval of convergence for a power series is to try using the ratio test. We recall that the ratio test tells us that if we let 𝐿 be equal to the limit as 𝑛 approaches ∞ of the absolute value of π‘Žπ‘› plus one divided by 𝑛, that is the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms. If this value of 𝐿 is less than one, then we can conclude the sum to ∞ of π‘Žπ‘› converges. And if 𝐿 is greater than one, then the sum to ∞ of π‘Žπ‘› diverges.

So let’s apply the ratio test to the power series given to us in the question. We have the limit as 𝑛 approaches ∞ of the absolute value of π‘Žπ‘› plus one divided by π‘Žπ‘› which is equal to the limit as 𝑛 approaches ∞ of our absolute value of our π‘Žπ‘› plus one term. Which is equal to π‘₯ to the power of three multiplied by 𝑛 plus one plus two all divided by five multiplied by 𝑛 plus one plus 10. And from this, we need to divide our term π‘Žπ‘›, which is π‘₯ to the power of three 𝑛 plus two all divided by five 𝑛 plus 10.

Now, let’s start expanding the parentheses in our first term. We get three multiplied by 𝑛 plus one plus two can be simplified to three 𝑛 plus five. And the denominator five multiplied by 𝑛 plus one plus 10 can be simplified to just five 𝑛 plus 15. And we recall that instead of dividing by our term π‘Žπ‘›, we can multiply it by the reciprocal of π‘Žπ‘›. We now recall one of our rules for exponents, which says that π‘₯ to the power of π‘Ž divided by π‘₯ to the power 𝑏 is just equal to π‘₯ to the power of π‘Ž minus 𝑏.

So we can simplify the shared factors of π‘₯ in the numerator and the denominator by just changing our numerator to have π‘₯ to the power of three 𝑛 plus five minus three 𝑛 minus two. Since three 𝑛 plus five minus three 𝑛 minus two is just equal to three, we have the limit as 𝑛 approaches ∞ of the absolute value of π‘₯ cubed multiplied by five 𝑛 plus 10 all divided by five 𝑛 plus 15. At this point, since we’re taking a limit as 𝑛 approaches ∞, we can notice that the value of π‘₯ is not dependent on the value of 𝑛. So π‘₯ is a constant with respect to 𝑛, and we can take the value of π‘₯ cubed out of our limit.

Well, we must be careful that it must be the absolute value of π‘₯ cubed since we’re taking the limit of an absolute value. This means the new limit that we have to evaluate is the limit as 𝑛 approaches ∞ of the absolute value of five 𝑛 plus 10 all divided by five 𝑛 plus 15. One way of evaluating limits like this is to divide both the numerator and the denominator by 𝑛. Evaluating this gives us the absolute value of π‘₯ cubed multiplied by the limit as 𝑛 approaches ∞ of the absolute value of five plus 10 over 𝑛 all divided by five plus 15 over 𝑛.

Since our limit is as 𝑛 is approaching ∞, both 10 divided by 𝑛 and 15 divided by 𝑛 are getting smaller and smaller, and so their limit is equal to zero. Therefore, when we evaluate this limit, we get the absolute value of five divided by five, which is just equal to one. And so our limit as 𝑛 approaches ∞ of the absolute value of π‘Žπ‘› plus one divided by π‘Žπ‘› is just equal to the absolute value of π‘₯ cubed. We now recall that our ratio test says that if this limit is less than one, then we can conclude our power series converges. And if this limit is greater than one, then we can conclude our power series diverges.

The first thing we notice is that the absolute value of π‘₯ cubed being less than one is exactly the same as saying that negative one is less than π‘₯ cubed is less than one. One way of finding the region of π‘₯ where this is true is to sketch a graph of 𝑦 equals π‘₯ cubed. We find that π‘₯ cubed is equal to one when π‘₯ is equal to one and π‘₯ cubed is equal to negative one when π‘₯ is equal to negative one. And we see that our region when our curve 𝑦 equals π‘₯ cubed is between one and negative one is when π‘₯ is between one and negative one. So what this means is that when π‘₯ is between negative one and one, we have that our limit 𝐿 is less than one, and so our power series must converge.

Similarly, from the sketch, we can see that if π‘₯ is bigger than one or if π‘₯ is less than negative one, then our output values of π‘₯ cubed are no longer between one and negative one. And so the absolute value of π‘₯ cubed must be bigger than one. Therefore, if π‘₯ is bigger than one, or if π‘₯ is less than negative one, then the size of our limit 𝐿 is bigger than one. And so by the ratio test, we have that our power series must be divergent. So far, what we have shown is that our interval of convergence for our power series must at least include the values from negative one to one. So this leaves us with the question, β€œAre one and negative one in our interval of convergence, which we will call 𝐼?”

Let’s go back to the series given to us in the question. Let’s start when π‘₯ is negative one. We see that our power series is equal to the sum from 𝑛 equals zero to ∞ of negative one to the power of three 𝑛 plus two all divided by five 𝑛 plus 10. We can simplify our numerator by noticing that negative one, always to the power of three 𝑛 plus two, is just equal to negative one all cubed all raised to the power of 𝑛 multiplied by negative one squared. We know that negative one squared is just equal to one. Similarly, negative one cubed can be evaluated to just be equal to negative one. Therefore, our numerator simplifies to negative one all raised to the 𝑛th power.

To evaluate this, we recall the alternating series test, which tells us that if a sequence π‘Žπ‘› is equal to negative one to the 𝑛th power multiplied by 𝑏𝑛 and 𝑏𝑛 isn’t always positive, decreasing sequence with the limit as 𝑛 approaches ∞ of 𝑏𝑛 equal to zero. Then we can conclude that the sum to ∞ of our sequence π‘Žπ‘› converges. So let’s set the sequence 𝑏𝑛 to be equal to one divided by five 𝑛 plus 10. First, let’s see if our sequence is positive. Well we know 𝑛 is always positive. So five 𝑛 plus 10 must always be positive, and then the reciprocal of this is also positive. So our sequence 𝑏𝑛 is always positive.

Similarly, we can see as we increase the value of 𝑛, what we’re doing is we’re making the denominator of our fraction bigger and bigger, which means that as we increase 𝑛, the fraction gets smaller and smaller. So our sequence 𝑏𝑛 is decreasing. Now, if we try to evaluate the limit as 𝑛 approaches ∞ of one divided by five 𝑛 plus 10, we see that as 𝑛 is getting bigger and bigger, the denominator of our fraction is getting bigger and bigger. And so we’re getting closer and closer to zero. So we know that the limit of 𝑏𝑛 as 𝑛 approaches ∞ is equal to zero. Therefore, what we’ve shown by alternating series test is the sum from 𝑛 equal zero to ∞ of negative one to the 𝑛th power divided by five 𝑛 plus 10 converges.

So in particular, what this tells us is that negative one must be in our interval of convergence. Now, let’s substitute π‘₯ is equal to one into our power series to determine whether one is in our interval of convergence. We get the sum from 𝑛 equals zero to ∞ of one raised to the power of three 𝑛 plus two all divided by five 𝑛 plus 10. One to the power of three 𝑛 plus two is just equal to one. So we get the sum from 𝑛 equals zero to ∞ of one divided by five 𝑛 plus 10. What we have is the sum of one divided by a linear factor. This is very similar to the harmonic series which we know diverges.

So let’s use a comparison test with the harmonic series. In fact, let’s use the limit comparison test, which tells us if we have two nonnegative sequences, π‘Žπ‘› and 𝑏𝑛, where the limit as 𝑛 approaches ∞ of π‘Žπ‘› divided by 𝑏𝑛 is positive. Then we can conclude that the sum to ∞ of π‘Žπ‘› and the sum to ∞ of 𝑏𝑛 either both converge or both diverge. So let’s set our first sequence to be equal to one over 𝑛 and our second sequence to be equal to one divided by five 𝑛 plus 10. This gives us the limit as 𝑛 approaches ∞ of π‘Žπ‘› divided by 𝑏𝑛 is equal to the limit as 𝑛 approaches of one over 𝑛 divided by one over five 𝑛 plus 10.

Instead of dividing by one over five 𝑛 plus 10, we can flip the fraction and multiply, giving us the limit as 𝑛 approaches ∞ of one over 𝑛 times five 𝑛 plus 10 divided by one. So now, we have to calculate the limit as 𝑛 approaches ∞ of five 𝑛 plus 10 all divided by 𝑛. We can evaluate our division by 𝑛 to give us the limit as 𝑛 approaches ∞ of five plus 10 divided by 𝑛. Since 𝑛 is approaching ∞, 10 divided by 𝑛 is approaching zero. So our limit is equal to the limit as 𝑛 approaches ∞ of the constant five, which is just equal to five. Therefore, since we have shown that we have two nonnegative sequences and the limit as 𝑛 approaches ∞ of π‘Žπ‘› over 𝑏𝑛 is positive, we can use the limit comparison test.

We know that the sum to ∞ of the harmonic series diverges. So we can conclude by our limit comparison test that the sum from 𝑛 equals zero to ∞ of one divided by five 𝑛 plus 10 must also diverge. What this means, in particular, is the value of π‘₯ is equal to one is not in our interval of convergence. So what we have shown is that the interval of convergence for the power series the sum from 𝑛 equals zero to ∞ of π‘₯ to the power of three 𝑛 plus two all divided by five 𝑛 plus 10 is equal to the interval starting at negative one and including negative one going all the way up to one, which it does not include.

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