Help! Convergence of Power Series, interval and radius of convergence question!

  • Thread starter knotman2
  • Start date
  • #1
9
0

Homework Statement


Determine the radius of convergence, the interval of convergence, and
the sum of the series
Summation from k=2 to ∞ of
k(x-2)^k+1.

Homework Equations


ratio test?


The Attempt at a Solution


possibly take the derrivitive of the power series, then find the sum then integrate?

use the ratio test to determine interval of convergence

(k+1) |x-2|^(k+2) / k|x-2|^(k+1) = [(k+1)/k] |x-2| ---> |x-2| as k ---> infinity.

Therefore the series converges for |x - 2|< 1 so 1 < x < 3 .
So the radius of convergence is 1
Is this linked to the sum? For example, if i think of taking the derivative of this function, you get f'(x)=sum from k=2 to infinity, of k(k+1)(x-2)^k which looks like the geometric series…But im not sure how to combine the ks? when 0<r<1, the geometric series converges to 1/1-r.
 
Last edited:

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766

Homework Statement


Determine the radius of convergence, the interval of convergence, and
the sum of the series
Summation from k=2 to ∞ of
k(x-2)^k+1.

Homework Equations


ratio test?


The Attempt at a Solution


possibly take the derrivitive of the power series, then find the sum then integrate?
So did you try the ratio test? What happened? Show us...
 
  • #3
9
0
have added now to the question above.
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766

Homework Statement


Determine the radius of convergence, the interval of convergence, and
the sum of the series
Summation from k=2 to ∞ of
k(x-2)^k+1.

Homework Equations


ratio test?


The Attempt at a Solution


possibly take the derrivitive of the power series, then find the sum then integrate?

use the ratio test to determine interval of convergence

(k+1) |x-2|^(k+2) / k|x-2|^(k+1) = [(k+1)/k] |x-2| ---> |x-2| as k ---> infinity.

Therefore the series converges for |x - 2|< 1 so 1 < x < 3 .
Good. Correct so far. You may need to check the end points.

So the radius of convergence is 1
Is this linked to the sum? For example, if i think of taking the derivative of this function, you get f'(x)=sum from k=2 to infinity, of k(k+1)(x-2)^k which looks like the geometric series…But im not sure how to combine the ks? when 0<r<1, the geometric series converges to 1/1-r.
You are thinking in the right direction. Try factoring (x-2)2 out of the sum and see if the remaining series looks like the derivative of something.
 
  • #5
9
0
I tried doing k(K+1)(x-2)^2(x-2)^k-2… this doesnt really look like the derivative of anything?
 
  • #6
9
0
Also how do i deal with the k(k+1)?
 
  • #7
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766
I tried doing k(K+1)(x-2)^2(x-2)^k-2… this doesnt really look like the derivative of anything?
Also how do i deal with the k(k+1)?
Your original series is

[tex]\sum_{k=2}^\infty k(x-2)^{k+1}[/tex]

What do you get when you factor (x-2)2 out of that sum? Just show me that.
 
  • #8
9
0
k(x-2)^2(x-2)^k-1
 
  • #9
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766
You left out the sum. Write it correctly with the (x-2)2 outside the sum.
 
  • #10
9
0
(x-2)^2 sum of from k=2 to infinity of k(x-2)^k-1
 
  • #11
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766
Click on this to see how to write it:

[tex](x-2)^2\sum_{k=2}^\infty k(x-2)^{k-1}[/tex]

The (x-2)2 is outside the sum so forget about it for now; just leave it there. You noticed the inside of the sum is the derivative of something. Write it that way using ' for derivative and remember for convergent power series the sum of the derivatives is the derivative of the sum.
 
  • #12
9
0
okay so
[tex]
(x-2)^2\(sum_{k=2}^\infty k(x-2)^{k-1})'
[/tex]
[tex]
(x-2)^2\(sum_{k=2}^\infty (x-2)^{k})
[/tex]
[tex]
(x-2)^2 (1/1-(x-2))'
[/tex]
[tex]
=(x-2)^2 (1/(x+1)^2?
[/tex]
 
  • #13
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766
okay so
[tex]
(x-2)^2\(sum_{k=2}^\infty k(x-2)^{k-1})'
[/tex]
[tex]
(x-2)^2\sum_{k=2}^\infty (x-2)^{k})
[/tex]
[tex]
(x-2)^2 (1/1-(x-2))'
[/tex]
[tex]
=(x-2)^2 (1/(x+1)^2?
[/tex]
You're getting close. Don't put anything between the \ and the sum. The tex gets a little tricky but you can learn by looking. Being careful with the derivatives, it looks like this:

[tex]
(x-2)^2\sum_{k=2}^\infty k(x-2)^{k-1}
[/tex]

[tex]
= (x-2)^2\sum_{k=2}^\infty ((x-2)^k)'
[/tex]

[tex]
= (x-2)^2\left(\sum_{k=2}^\infty (x-2)^k\right)'
[/tex]

But be careful with the sum inside the parentheses. It doesn't start at k = 0 and the first term isn't 1. Once you have it, you can differentiate it and you are on your way
 
  • #14
9
0
[tex]

= (x-2)^2\left(\sum_{k=2}^\infty (x-2)^k\right)'

[/tex]
[tex]

= (x-2)^2\((1/1-(x-2))-1-(x-2))'

[/tex]
[tex]

= (x-2)^2\((1/(x+1)^2)-1)

[/tex]
?
 
  • #15
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766
[tex]

= (x-2)^2\left(\sum_{k=2}^\infty (x-2)^k\right)'

[/tex]
[tex]

= (x-2)^2\((1/1-(x-2))-1-(x-2))'

[/tex]
[tex]

= (x-2)^2\((1/(x+1)^2)-1)

[/tex]
?
That last step has mistakes but it is the right idea. Remove some parentheses and collect some terms before you take the derivative and watch your signs. You should be able to take it from here.
 

Related Threads on Help! Convergence of Power Series, interval and radius of convergence question!

Replies
2
Views
2K
Replies
10
Views
2K
Replies
2
Views
1K
Replies
6
Views
10K
Replies
8
Views
2K
Replies
1
Views
818
Replies
1
Views
1K
Top