Master Power Series Convergence with Expert Help - Examples Included

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SUMMARY

This discussion focuses on the convergence of three power series using the ratio test and alternating series test. The first series, ∑[k2(x-2)k]/[3k], has an interval of convergence I = (1,3). The second series, ∑[(x-4)n]/[(n)(-9)n], is incorrectly assessed with an interval of I = (-13,5), while the correct interval is centered at x=4, likely I = (3,5). The third series, ∑[2k(x-3)k]/[k(k+1)], has endpoints I = [5/2,7/2) but diverges at x=5/2 as the terms do not approach zero.

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PhysicsCollegeGirl
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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'hospital's, but it always shows me that it diverges.
 
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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.
 
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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