Assume that the series ∑ aₙ(x − 2)ⁿ converges for x = −1 and d...

Question

Assume that the series ∑ aₙ(x − 2)ⁿ converges for x = −1 and diverges for x = 6. Answer true (T), false (F), or not enough information given (N) for the following statements about the series.
f. Diverges for x = 4.9

f. Diverges for x = 4.9

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. True or false, if the power series, the sum of a n multiplied by parentheses of x minus 3 to the power of n converges at x equals 0 and diverges at x equals 9, then it also diverges at x equals 5.9. OK, so it appears for this particular problem we're given a statement and we're asked to determine whether or not this statement will be A true or B false. So now that we know what we're ultimately trying to solve for, we need to recall a note that for a power series, which states that the sum of A N multiplied by parentheses of x minus c to the power of n. Is centered at. C, that will mean there is a radius of convergence, capital R, that is greater than or equal to 0. Fantastic. So for such a series, we need to recall and note the following. So once again, so for a power series that is centered at C, we have our radius of convergence. R is greater than or equal to 0, and for this particular series, it will converge for all x such that the absolute value of x minus c will be less than r. And we will note that it diverges for all x such that the absolute value of x minus c is greater than r, and the behavior at the absolute value of x minus c must equal r, and this must be checked separately. So the behavior specifically at the absolute value of X minus C is equal to R must be checked separately. So with that said. We now need to consider the convergence at x is equal to 0. So we need to note that R is greater than or equal to the absolute value of 0 minus 3, which is equal to 3. That will mean therefore that the smallest possible r value will be that our men. Will be equal to 3. And now we need to consider divergence at, so we had convergence earlier, so now we need to consider divergence at x equals 9. So I put a C in quotes for convergence and a D in for divergence. So considering divergence at x equals 9, we can state that R is less than the absolute value of 9 minus 3, which is equal to 6. And if R is greater than 6, that will mean that the series would converge at X equals 9. Thus, we could write that R is less than or equal to 6. Thus, as a result, the largest possible R value will be. Our max and our max will be simply equal to 6. Fantastic. So moving right along here. So we can safely say the following now based on what we've learned so far. So we need to note that when we combine these bounds, this will give us that R is an element of the closed interval square brackets of 3.6, and this represents the maximum possible interval of convergence. We can also go ahead to say that if the absolute value of x minus 3 is less than 3, then the absolute value of x minus 3 is less than r, for every r is an element of the closed interval 3.6. The convergence at such an x is guaranteed. And if the absolute value of x minus 3 is greater than 6, then the absolute value of x minus 3 is greater than r, for every r is an element of the closed interval 3.6, and the divergence at such an x is guaranteed. And finally, if 3 is less than or equal to the absolute value of x minus 3, which is less than or equal to 6, nothing definite can be concluded from the given information alone. Whether the series converges there depends on the actual r within our closed interval square brackets of 3.6. And on the point-wise boundary behavior. So with that in mind, we now need to apply this to. X equals 5.9, so we need to take the absolute value of 5. 0.9 minus 3. Which will be equal to 2.9, which is less than 3, and we need to note that 3 is less than or equal to R, then that will mean that 2.9 is less than R. And due to this fact, thus, the series converges absolutely at, so once again, our series will converge absolutely, so I put CA in quotes, so it's converging absolutely at X equals 5.9, and because it is converging absolutely at X equals 5.9, we could safely say that our final answer will be B false. And that's it. That is our final answer. We have determined that the statement made by the prom itself is false. Hooray, we did it. So looking at our multiple choice answers, our final answer without a doubt, once again will be multiple choice answer letter B, false. And that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Assume that the series ∑ aₙ(x − 2)ⁿ converges for x = −1 and diverges for x = 6. Answer true (T), false (F), or not enough information given (N) for the following statements about the series.f. Diverges for x = 4.9

Watch next

Assume that the series ∑ aₙ(x − 2)ⁿ converges for x = −1 and diverges for x = 6. Answer true (T), false (F), or not enough information given (N) for the following statements about the series.
f. Diverges for x = 4.9