Video: Finding the Interval of Convergence of a Power Series

Find the interval of convergence for the power series βˆ‘_(𝑛 = 1) ^(∞) (π‘₯ + 3)^(𝑛)/(6𝑛).

06:43

Video Transcript

Find the interval of convergence for the power series the sum from 𝑛 equals one to ∞ of π‘₯ plus three to the 𝑛th power divided by six 𝑛.

The question wants us to find the interval of convergence for this power series. And we recall that an interval of convergence for a power series is an interval which contains all of the values of π‘₯ such that the series converges. To help us find the interval of convergence of this power series, we’re going to use the ratio test. And we recall the ratio test tells us if we have an infinite series the sum of π‘Ž 𝑛, where the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms is less than one. Then our series converges absolutely. However, if our series the sum of π‘Ž 𝑛 has the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms is greater than one, then the ratio test tells us that our series must be divergent.

In our case, our series is the sum from 𝑛 equals one to ∞ of π‘₯ plus three to the 𝑛th power divided by six 𝑛. So we’ll set our sequence π‘Ž 𝑛 equal to this and then attempt to use the ratio test. Using this gives us the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms is equal to the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one divided by π‘Ž 𝑛. Where π‘Ž 𝑛 plus one is equal to π‘₯ plus three to the power of 𝑛 plus one divided by six times 𝑛 plus one. And π‘Ž 𝑛 is equal to π‘₯ plus three to the 𝑛th power divided by six 𝑛.

We can simplify this limit. Instead of dividing by the fraction π‘₯ plus three to the 𝑛th power divided by six 𝑛, we can multiply by the reciprocal of this fraction. This gives us the limit as 𝑛 approaches ∞ of the absolute value of π‘₯ plus three to the power of 𝑛 plus one divided by six times 𝑛 plus one multiplied by six 𝑛 divided by π‘₯ plus three to the 𝑛th power. We can simplify this limit by canceling the shared factor of six in our numerator and our denominator and canceling 𝑛 of the shared factors of π‘₯ plus three in our numerator and our denominator. This leaves us with the limit as 𝑛 approaches ∞ of the absolute value of π‘₯ plus three times 𝑛 divided by 𝑛 plus one.

Since our limit is as 𝑛 is approaching ∞, we can see that π‘₯ plus three remains constant. It’s invariant with respect to 𝑛. So we can take this constant outside of our limit, where we must be careful to take the absolute value of π‘₯ plus three since we were calculating the limit of an absolute value. We now need to evaluate the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 divided by 𝑛 plus one. There’s a few different ways of doing this. For example, we could divide 𝑛 by 𝑛 plus one by using algebraic division. However, we’re going to use the method of dividing both our numerator and our denominator by the highest power of 𝑛 which appears in the fraction.

We can see the only power of 𝑛 which appears in our fraction is 𝑛 itself, so we’ll divide both our numerator and our denominator by 𝑛. Dividing the numerator of 𝑛 by 𝑛 just gives us one. And dividing our denominator of 𝑛 plus one through by 𝑛 will give us one plus one divided by 𝑛. So this gives us the absolute value of π‘₯ plus three multiplied by the limit as 𝑛 approaches ∞ of the absolute value of one divided by one plus one over 𝑛.

We’re now ready to evaluate this limit. Since our limit is as 𝑛 is approaching ∞, the one in our numerator and the one in our denominator remain constant. However, as 𝑛 is approaching ∞, one divided by 𝑛 is approaching zero. This means our limit is approaching the absolute value of one divided by one, which is just equal to one. Therefore, we’ve shown the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms of the series given to us in the question is equal to the absolute value of π‘₯ plus three.

And remember the ratio test tells us if this value is less than one, then our series is convergent. And if this value is greater than one, then the series is divergent. So the ratio test tells us that our series will converge if the absolute value of π‘₯ plus three is less than one. And saying the absolute value of π‘₯ plus three is less than one is the same as saying that π‘₯ plus three is greater than negative one and π‘₯ plus three is less than one. We can then subtract three from all parts of this inequality to see that π‘₯ must be bigger than negative four and π‘₯ must be less than negative two.

Similarly, using the ratio test, we can conclude if the absolute value of π‘₯ plus three is greater than one, then our series will be divergent. And using a similar argument, we can show that this is the same as saying π‘₯ is less than negative four or π‘₯ is greater than negative two. But remember, the interval of convergence contains all of the values of π‘₯ such that our series is convergent. And we’ve looked at all of the values of π‘₯ except when π‘₯ is equal to negative four and when π‘₯ is equal to negative two since at these two values of π‘₯, the ratio test was inconclusive. We found the limit as 𝑛 approached ∞ of the absolute value of the ratio of successive terms for these two values of π‘₯ was just equal to one.

To help us determine whether the series converges or diverges in these two cases, we’re going to clear some space and then substitute the values of π‘₯ is equal to negative four and π‘₯ is equal to negative two directly into our power series. We’ll start when π‘₯ is equal to negative two. This gives us the sum from 𝑛 equals one to ∞ of negative two plus three to the 𝑛th power divided by six 𝑛.

We can start by taking the constant factor of one-sixth outside of our sum. Next, we can evaluate negative two plus three to just give us one. And we know that one to the 𝑛th power is always equal to one. And we’ve now shown that our series is equal to one-sixth multiplied by the harmonic series, which we know diverges. Therefore, our series was divergent when π‘₯ was equal to negative two. And so this won’t be in our interval of convergence.

We can do the same when π‘₯ is equal to negative four. We can simplify our series to give us one-sixth multiplied by the sum from 𝑛 equals one to ∞ of negative one to the 𝑛th power divided by 𝑛. We can now see that our summand negative one to the 𝑛th power divided by 𝑛 is an alternating sequence. And one over 𝑛 is positive and decreasing. Therefore, by the alternating series test, we know that this series is convergent. Therefore, we’ve shown all of the values of π‘₯ where our series is convergent is when π‘₯ is greater than negative four and less than negative two and when π‘₯ is equal to negative four.

Therefore, we’ve shown that the interval of convergence for the power series the sum from 𝑛 equals one to ∞ of π‘₯ plus three to the 𝑛th power divided by six 𝑛 is when π‘₯ is greater than or equal to negative four and when π‘₯ is less than negative two.

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