Someone please help me with this seemingly geometric series

**s3a** · May 4th 2010, 10:07 AM

Question:
To show that the following
http://www.wolframalpha.com/input/?i=sum+of+(k*(2%2F3)^k)+from+1+to+inf
is convergent conditionally or absolutely.

What I am having trouble with is the fact that it is multiplied by k. I realize that exponentiating something to k will grow faster than multiplying it by k but that is not sufficient at all for me to prove that this series converges. Actually I just used the ratio test now and found 2/3 which converges. So now a new question arises; is the ratio test valid for geometric series? Is this series a geometric one even though it is multiplied by k?

Any input would be greatly appreciated!
Thanks in advance!

**Failure** · May 4th 2010, 10:19 AM

Originally Posted by s3a

Question:
To show that the following
http://www.wolframalpha.com/input/?i=sum+of+(k*(2%2F3)^k)+from+1+to+inf
is convergent conditionally or absolutely.

What I am having trouble with is the fact that it is multiplied by k. I realize that exponentiating something to k will grow faster than multiplying it by k but that is not sufficient at all for me to prove that this series converges. Actually I just used the ratio test now and found 2/3 which converges. So now a new question arises; is the ratio test valid for geometric series? Is this series a geometric one even though it is multiplied by k?

Any input would be greatly appreciated!
Thanks in advance!

You may want to first derive a closed form for the series $f(x) := \sum_{k=1}^\infty k x^k$

This can be done like this
$f(x)=x\cdot\sum_{k=1}^\infty k x^{k-1}=x\cdot\sum_{k=0}^\infty \frac{d}{dx} x^k$

$=x\cdot \frac{d}{dx} \sum_{k=0}^\infty x^k=x\cdot \frac{d}{dx}\frac{1}{1-x}=\frac{x}{(1-x)^2}$
Now just plug in $2/3$ for x, to get $f(2/3)=\tfrac{2}{3}\cdot\frac{1}{\left(1-\frac{2}{3}\right)^2}=6$ .

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