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

[knotman2]

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 I am not sure how to combine the ks? when 0<r<1, the geometric series converges to 1/1-r.

 

[LCKurtz]

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...

 

[knotman2]

have added now to the question above.

 

[LCKurtz]

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 I am 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.

 

[knotman2]

I tried doing k(K+1)(x-2)^2(x-2)^k-2… this doesn't really look like the derivative of anything?

 

[knotman2]

Also how do i deal with the k(k+1)?

 

[LCKurtz]

I tried doing k(K+1)(x-2)^2(x-2)^k-2… this doesn't really look like the derivative of anything?

Also how do i deal with the k(k+1)?

Your original series is

\sum_{k=2}^\infty k(x-2)^{k+1}

What do you get when you factor (x-2)² out of that sum? Just show me that.

 

[knotman2]

k(x-2)^2(x-2)^k-1

 

[LCKurtz]

You left out the sum. Write it correctly with the (x-2)² outside the sum.

 

[knotman2]

(x-2)^2 sum of from k=2 to infinity of k(x-2)^k-1

 

[LCKurtz]

Click on this to see how to write it:

(x-2)^2\sum_{k=2}^\infty k(x-2)^{k-1}

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.

 

[knotman2]

okay so
<br /> (x-2)^2\(sum_{k=2}^\infty k(x-2)^{k-1})&#039;<br />
<br /> (x-2)^2\(sum_{k=2}^\infty (x-2)^{k})<br />
<br /> (x-2)^2 (1/1-(x-2))&#039;<br />
<br /> =(x-2)^2 (1/(x+1)^2?<br />

 

[LCKurtz]

okay so
<br /> (x-2)^2\(sum_{k=2}^\infty k(x-2)^{k-1})&#039;<br />
<br /> (x-2)^2\sum_{k=2}^\infty (x-2)^{k})<br />
<br /> (x-2)^2 (1/1-(x-2))&#039;<br />
<br /> =(x-2)^2 (1/(x+1)^2?<br />

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:

<br /> (x-2)^2\sum_{k=2}^\infty k(x-2)^{k-1}<br />

<br /> = (x-2)^2\sum_{k=2}^\infty ((x-2)^k)&#039;<br />

<br /> = (x-2)^2\left(\sum_{k=2}^\infty (x-2)^k\right)&#039;<br />

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

 

[knotman2]

<br /> <br /> = (x-2)^2\left(\sum_{k=2}^\infty (x-2)^k\right)&#039;<br /> <br />
<br /> <br /> = (x-2)^2\((1/1-(x-2))-1-(x-2))&#039;<br /> <br />
<br /> <br /> = (x-2)^2\((1/(x+1)^2)-1)<br /> <br />
?

 

[LCKurtz]

<br /> <br /> = (x-2)^2\left(\sum_{k=2}^\infty (x-2)^k\right)&#039;<br /> <br />
<br /> <br /> = (x-2)^2\((1/1-(x-2))-1-(x-2))&#039;<br /> <br />
<br /> <br /> = (x-2)^2\((1/(x+1)^2)-1)<br /> <br />
?

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.