# Thread: Geometric Sum Question: sum [k=1, infty] [k * q^{k-1} * p]

1. ## Geometric Sum Question: sum [k=1, infty] [k * q^{k-1} * p]

Can someone show me how to solve this equation?

S = Sum of K*p(x=k)

Where K starts and 1 and approaches infinity
And p(x=k) = (1-pi/4)^(k-1)*pi/4

picture of the problem here

The instructor gave the answer 4/pi but I cannot seem to reproduce the answer.

Any help will be greatly appreciated. Thank you very much!

2. Well, you say you can't reproduce the answer, which directly implies you've tried at least one method. That's excellent! Trying things is usually the best way to learn. Unfortunately, we cannot provide any feedback to you without seeing these efforts. For all we know, you could be nearly to the correct solution but have made a tiny, easily correctable error along the way. So no sense in us starting over at square one. Please re-read the rules of this forum, as laid out in the Read Before Posting thread that's stickied at the top of each subforum. Then please share with us all of the work you've done on this problem, even the parts you know for sure are wrong. Thank you.

3. Originally Posted by ksdhart2
Well, you say you can't reproduce the answer, which directly implies you've tried at least one method. That's excellent! Trying things is usually the best way to learn. Unfortunately, we cannot provide any feedback to you without seeing these efforts. For all we know, you could be nearly to the correct solution but have made a tiny, easily correctable error along the way. So no sense in us starting over at square one. Please re-read the rules of this forum, as laid out in the Read Before Posting thread that's stickied at the top of each subforum. Then please share with us all of the work you've done on this problem, even the parts you know for sure are wrong. Thank you.
Oh sorry! Here is my work https://ibb.co/cZ8TNn. On the right side I was able get it to a geometric sum format and tried to use the formulat s=a1/(1-r)

Then on the left side I tried to use the formula S = 1/(1-r) Both of which failed to produce the desire result. Sorry of the mess.

4. Oh, hey, sorry about the delay. For some reason I didn't get notified of your response. If I hadn't happened to manually check out of curiosity, it may have gone forever unnoticed. I'll have to double check my notification settings. Anyway... what you've done so far is good. You've successfully demonstrated that part of the problem involves a geometric series. More specifically, what you have there is the formula for the expected value of a geometric distribution, which models the probability of getting a "success" on the kth trial. Just for convenience sake, let's define a few values. Let $p = \dfrac{\pi}{4}$ and let $q = 1 - p = \dfrac{3\pi}{4}$. We then have:

$\displaystyle S = \sum_{k=1}^{\infty} k \cdot q^{k-1} \cdot p = pq + 2pq^2 + 3pq^3 + \dots$

If we multiply this through by q we get:

$\displaystyle q \cdot S = \sum_{k=1}^{\infty} k \cdot q^{k} \cdot p = pq^2 + 2pq^3 + 3pq^4 + \dots$

And hence:

$\displaystyle S - q \cdot S = pq + (2pq^2 - pq^2) + (3pq^3 - 2pq^3) + (4pq^4 - 3pq^4) + \dots$

5. Originally Posted by ksdhart2
Oh, hey, sorry about the delay. For some reason I didn't get notified of your response. If I hadn't happened to manually check out of curiosity, it may have gone forever unnoticed. I'll have to double check my notification settings. Anyway... what you've done so far is good. You've successfully demonstrated that part of the problem involves a geometric series. More specifically, what you have there is the formula for the expected value of a geometric distribution, which models the probability of getting a "success" on the kth trial. Just for convenience sake, let's define a few values. Let $p = \dfrac{\pi}{4}$ and let $q = 1 - p = \dfrac{3\pi}{4}$. We then have:

$\displaystyle S = \sum_{k=1}^{\infty} k \cdot q^{k-1} \cdot p = pq + 2pq^2 + 3pq^3 + \dots$

If we multiply this through by q we get:

$\displaystyle q \cdot S = \sum_{k=1}^{\infty} k \cdot q^{k} \cdot p = pq^2 + 2pq^3 + 3pq^4 + \dots$

And hence:

$\displaystyle S - q \cdot S = pq + (2pq^2 - pq^2) + (3pq^3 - 2pq^3) + (4pq^4 - 3pq^4) + \dots$

Ah thank you very much!
This gives me:
$\displaystyle S(1-q)=p(q+q^2+q^3+\dots)$
Then
$\displaystyle S=\dfrac{p}{(1-q)}*\dfrac{1}{(1-q)}$
And finally
$\displaystyle S=\dfrac{p}{p}*\dfrac{1}{p}=\dfrac{4}{\pi}$

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