Intervals of Convergence
In Exercises 1–36, for what val...

Question

Intervals of Convergence

In Exercises 1–36, for what values of x does the series converge (c) conditionally?

∑ (from n = 1 to ∞) [ (3x + 1)^(n + 1) / (2n + 2) ]

Accepted Answer

Hello. In this video, we are going to be determining the values of X where the series converges conditionally. So, the series that is given to us is the infinite series. From 1 to infinity of 2 x plus 3 raise the power of n divided by 4 n plus 4. Now, in order to determine condition or condition or absolute convergence, we want to apply the ratio test. Now here, we are going to name the function of our summation as a subn. The ratio test. Is a test with the infinite limit, a limit as N goes to infinity, of the absolute value of A_N + 1 divided by AN. So in order to set up this limit, we need to plug in the value m + 1 into our summation, then divide by the original summation. So by applying the ratio test, we can set up the limit as the following. We will have the limit as n approaches infinity of the absolute value of 2 x + 3 raised to the power of n + 1 divided by 4 multiplied by n + 1 + 4. Now here we could make a complex fraction and divide by the original summation, but equivalently we can multiply by the original summations reciprocal. So, by multiplying by the reciprocal, we will have the product of the AM + 1 argument multiplied by 4 N + 4, divided by 2 x + 3 raised the power of N. So there's a bit of simplification that we can do here. First, if we simplify 2 x plus 3 quantities, we can group them up as 2 x + 3, raise the power of n + 1, divided by 2 x + 3, raise the power of n. Now by our exponential properties, we can rewrite the numerator as a product of exponential values. We can rewrite the numerators 2 x plus 3, raise the power of n multiplied by 2 x plus 3, and this will be divided by 2 x plus 3, raise the power of n. Notice that 2 x + 3 raised to the power of n will simplify out of the numerator and denominator. This will leave us with just the quantity 2 x + 3. Now let's take a look at the 4 + 4 quantities. In the numerator, we will have 4 n + 4, and in the denominator, we will have 4 multiplied by n + 1 + 4. Now simplifying the numerator will allow us to rewrite this fraction as 4 n + 4 divided by 4 n + 8. The only thing we can really do here is factor out a 4 from the numerator and denominator, and that will allow us to rewrite this quantity as n + 1 divided by M + 2. So if we place back the simplifications into our limit, we will have the limit as n approaches infinity of the absolute value. Of the quantity 2 x + 3 multiplied by the quantity m + 1. Divided by m + 2. Now here 2 x plus 3 is acting as a constant to the limit. If we move this to the front of the limit with respect to the absolute value, we will have the absolute value of 2 x plus 3 multiplied by the limit as n approaches infinity of the absolute value of m + 1 divided by m plus 2. Now through our infinite limit properties, this is going to simplify to be 1, and we are left with the absolute value of 2 x plus 3. Now, if we want to determine the values of conditional convergence, what we want to do is we want to set this limit. To the condition that it will converge to, so we want this value to be less than one with respect to the ratio test. Now, this tells us that we have a radius of convergence of one. And if we rewrite this inequality to be -1 less than 2 x + 31 and we were to isolate for x, we can rewrite this interval to be x between the values of -2 and -1. But now what we need to do is we need to go ahead and determine. The points of conditional convergence. So what we're going to do is we're going to take the end points of our integral of convergence and plug them back into the original summation. Starting with x is equal to -2, plugging this back into the original summation leaves us with the summation from 1 to infinity. Of -1 to the power of n divided by 4 n plus 4. Now here we have an alternating series, but if we were to use the alternating series test. The result of this test is going to give us the limit. As n approaches infinity of the absolute value 1 divided by 4 plus 4, which is equal to 0, so this is going to converge conditionally by the alternating series test. So we know that we have conditional convergence that x is equal to -2. Now we will go ahead and test the value of x is equal to -1. When x is equal to -1, we have the infinite summation from 1 to infinity of 1 divided by 4 n plus 4. Now this is going to be a divergent harmonic series, which means that we have no convergence at the end point of X is equal to -1. So with that being said, there's only going to be conditional convergence at one end point of the interval of convergence, and that is going to be at X equal to -2, which is the solution to the problem. And the overall solution is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Intervals of ConvergenceIn Exercises 1–36, for what values of x does the series converge (c) conditionally?∑ (from n = 1 to ∞) [ (3x + 1)^(n + 1) / (2n + 2) ]

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Intervals of Convergence
In Exercises 1–36, for what values of x does the series converge (c) conditionally?
∑ (from n = 1 to ∞) [ (3x + 1)^(n + 1) / (2n + 2) ]