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 ∞) [ (√(n + 1) − √n)(x − 3)ⁿ ]

Accepted Answer

Hello. In this video, we are going to determine the values of X where the given series converges conditionally. So, the series given to us is X minus 5 raised to the power of N multiplied by the quantity the square root of N minus 1 minus the square root of N minus 2. Now, there is a test that allows us to find the interval of convergence that will tell us the values of X where the series converges conditionally. That test is called the ratio test. For the ratio test, we are going to name the function of the series a subn. And in order to find the interval of convergence, we need to take the limit as n approaches infinity of the absolute value a subn + 1 divided by a subn. So we are going to plug in the quantity and plus 1 into our summation, and then divide by the original summation. And in order for the series to converge, we want the limit L to be less than 1. So let's go ahead and start this problem by setting up the infinite limit. So we will have the limit, as n approaches infinity of the absolute value x minus 5 raised to the power of n + 1 multiplied by the quantity square root n + 1 minus 1, which is just n, minus the square root of n + 1 minus 2, which is n minus 1. Then this will be divided by the original series, which is going to be X minus 5, raised to the power of n, multiplied by the square root N minus 1 minus the square root of N minus 2. Now, in order to approach this type of limit, we are going to first simplify the x minus 5 quantities. Taking a look at the x minus 5 quantities in the numerator, we have X minus 5 raised to the power of n + 1, divided by x minus 5 raised to the power of n. Now by using our exponential properties, we can rewrite the numerator to be X minus 5 raised to the power of N multiplied by X minus 5, and this will be divided by X minus 5 raised to the power of N. Now the power of n quantities will simplify out of the numerator and denominator, leaving us with x minus 5. So we are left with the limit as n approaches infinity of the absolute value x minus 5 multiplied by the square root of n minus the square root of n minus 1 divided by the square root of n minus 1 minus the square root of n minus 2. Now here, X minus 5 is acting as a constant to the sum to the limit. So what we can go ahead and do is move this to the front of the limit as a constant value. That will allow us to rewrite the limit as the absolute value of x minus 5 multiplied by the limit as n approaches infinity of the absolute value of the square root n minus the square root n minus 1 divided by the square root of n minus 1 minus the square root of n minus 2. Now if we were to apply the infinite limit, the infinite limit will simplify to be 0 divided by 0, which is an indeterminate form. Now, one way we can rewrite this infinite limit is to rationalize the numerator and denominator. Now, if we were to rationalize the numerator and denominator, we can further rewrite the limit to be the absolute value of Xx minus 5 multiplied by the limit as n approaches infinity of the absolute value square root n minus 1 plus the square root of n minus 2, divided by the square root of n plus the square root of n minus 1. And if we were to apply the infinite limit properties, this is going to simplify to be 1, so the overall limit is going to simplify to be the absolute value of x minus 5. Now we want this quantity to cause our summation to be convergent, so we need X minus 5 to be less than 1. And this tells us that we have a radius of convergence of one. Now, we need to find the values of x which act as our endpoints for the interval of convergence. In order to do this, we are going to rewrite the inequality as x minus 5 between the values of -1 and 1. Now, by adding 5 to both sides of the inequality in order to isolate for X, we will end up with the values X is between 4 and 6. So these are going to be the values of the interval of convergence. Now, the first thing we want to do is we want to check to see if we have any absolute convergence in the summation. So, we need to do this by observing the absolute value of the original summation with 4 and 6 plugged in. Now, if we take the absolute value, 4 and 6 are going to produce the same quantity for X. -5 raises the power of n. Plugging in x is equal to 4 and x is equal to 6 will leave us with positive and negative 1 to the power of n. But the moment that we take the absolute value of this type of quantity, it is going to eliminate the alternating term. So by taking the absolute value of our original summation, we will have the summation from n is equal to infinity of the square root. Of n minus 1. Minus the square root of n minus 2. Now here we can go ahead and rewrite the summation by taking or by rationalizing the summation, and by rationalizing the summation, we will get the infinite series from 2 to infinity of 1 divided by the square root of n minus 1 plus the square root of n minus 2. Now, by the direct comparison test, this is going to be a divergent series, and since this is a divergent series, this tells us that there is no absolute convergence on any endpoint of the interval of convergence. But now we want to check for conditional convergence. In order to check for conditional convergence, we are going to plug in the endpoints 4 and 6 back into the original series and determine the convergence. So starting at X is equal to 4, plugging this into the original series will give us the infinite summation from 2 to infinity. Of -1 raised to the power of n, and we are going to work with the rationalized form of the of the summation. So this will be -1 to the power of n multiplied by 1 divided by the square root of n minus 1 plus the square root of n minus 2. Here we have an alternating series, but this is going to converge by the alternating series test. So we have conditional convergence at x is equal to 4. And if we repeat this process for x is equal to 5, we get -1 to the power of n, which is just 1. So this is going to be the infinite summation from 2 to infinity of 1 divided by the square root of n minus 1 plus the square root of n minus 2. And just like we've seen above, this is going to be a divergent series by the direct comparison test. So that means that we have no convergence at all and x is equal to 5. That means there is only one value of x where we have conditional convergence, and that is going to be at x is equal to 4, which is the solution to the problem. And the overall solution to the problem 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 ∞) [ (√(n + 1) − √n)(x − 3)ⁿ ]

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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 ∞) [ (√(n + 1) − √n)(x − 3)ⁿ ]