Master Power Series Convergence with Expert Help - Examples Included

In summary: I = (-13,5)The interval of convergence of this power series is centered at x=4. 3. ∑[2k(x-3)k]/[k(k+1)]I = [5/2,7/2)You can tell immediately that this is wrong. The interval of convergence of this power series is centered at x=4. However, in the third problem, x=5/2, the power series seems to go to zero, but the alternating series test says it doesn't
  • #1
PhysicsCollegeGirl
6
2

Homework Statement


[/B]
There are three problems that I am struggling with.

1. ∑[k2(x-2)k]/[3k]
2. ∑[(x-4)n]/[(n)(-9)n]
3. ∑[2k(x-3)k]/[k(k+1)]

The Attempt at a Solution



On the first two I am having problems finding the end-points of the interval of convergence. I use the ratio test.

1. ∑[k2(x-2)k]/[3k]
Here I get I = (1,3)

2. ∑[(x-4)n]/[(n)(-9)n]
I = (-13,5)

And in the last one I got the end-points but I am not sure why the left one would diverge.
It says it diverges in the answers but I cannot see it.

3. ∑[2k(x-3)k]/[k(k+1)]

I = [5/2,7/2)

I use the alternating series test to evaluate x=5/2 and it doesn't seem to go to zero. I do l'hopitals, but it always shows me that it diverges.
 
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  • #2
PhysicsCollegeGirl said:

Homework Statement


[/B]
There are three problems that I am struggling with.

1. ∑[k2(x-2)k]/[3k]
2. ∑[(x-4)n]/[(n)(-9)n]
3. ∑[2k(x-3)k]/[k(k+1)]

The Attempt at a Solution



On the first two I am having problems finding the end-points of the interval of convergence. I use the ratio test.

1. ∑[k2(x-2)k]/[3k]
Here I get I = (1,3)

2. ∑[(x-4)n]/[(n)(-9)n]
I = (-13,5)
You can tell immediately that this is wrong. The interval of convergence of this power series is centered at x=4. I could believe (3,5) but not (-13,5). (I don't know about the endpoints.)
And in the last one I got the end-points but I am not sure why the left one would diverge.
It says it diverges in the answers but I cannot see it.

3. ∑[2k(x-3)k]/[k(k+1)]

I = [5/2,7/2)

I use the alternating series test to evaluate x=5/2 and it doesn't seem to go to zero.
Look at this again. The terms are (-1)k/[k(k+1)] at x=5/2.
 
  • #3
PhysicsCollegeGirl said:
1. ∑[k2(x-2)k]/[3k]
Here I get I = (1,3)

It would be really helpful if you showed your work. Simply writing down the wrong answer doesn't give us much to go on to figure out where your mistakes are.
 
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Related to Master Power Series Convergence with Expert Help - Examples Included

1. What is a power series?

A power series is a series of the form ∑n=0∞ an(x-c)n, where an is a sequence of real or complex numbers and c is a constant. It is a type of infinite series that can be used to represent functions and can be used in calculus and other branches of mathematics.

2. How do you find the radius of convergence for a power series?

The radius of convergence for a power series can be found by using the ratio test. The ratio test states that if limn→∞ |an+1/an| = L, then the series converges if L < 1 and diverges if L > 1. The radius of convergence is equal to 1/L.

3. What is the interval of convergence for a power series?

The interval of convergence for a power series is the set of all values of x for which the series converges. It can be found by using the ratio test and determining the values of x that satisfy the condition L < 1. The interval of convergence can be an open interval, a closed interval, or a combination of both.

4. Can a power series represent any function?

No, a power series can only represent certain types of functions. The function must be analytic, meaning it can be represented by a power series. Examples of analytic functions include polynomials, trigonometric functions, and exponential functions. However, not all functions are analytic and therefore cannot be represented by a power series.

5. How do you use a power series to approximate a function?

To approximate a function using a power series, you can use the Maclaurin series, which is a special case of a power series where c = 0. The Maclaurin series can be used to approximate a function at a specific point by plugging in the value of x into the series. The more terms you include in the series, the more accurate the approximation will be.

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