# Is there a way to determine the summation of all "even" elements of an infinite geometric sequence?*

*Algebraically and not with by adding them for the first 100 or whatever and estimating

This came up when I came across a question in my probability class where I was asked about a game where 2 players take turns to accomplish a task until one succeeds. Each attempt has a probability of $p = x$ for success (I'm using $x$ so it's general, but in my case, I got .35) This makes a geometric series of $x(1-x)^{(n-1)}$ for $n$ being a positive integer where $n$ equals nth attempt. The question asked was what's the probability the first player wins, I wrote it as: $$\lim_{t\to \infty} \sum_{k=0}^{t}x*(x-1)^{2k}$$

Is there any way to solve this algebraically or something else or do I have to rely on computers for this task?

Sorry for this awful formatting I'm asking this on a phone and I can't post images since I'm new and I can't write latex on mobile.

Even terms of a geometric progression still form a geometric progression!

So, using the formula $$\sum\limits_{k=0}^{\infty} p^k = \frac{1}{1-p}$$

we ca easily derive

$$\sum\limits_{k=0}^{\infty} p^{2k} = \sum\limits_{k=0}^{\infty} (p^2)^k = \frac{1}{1-p^2}$$

I hope you can go from here.

$$\sum_{k=0}^\infty x(x-1)^{2k} = x\cdot\sum_{k=0}^\infty\left((x-1)^2\right)^k$$

is again a geometric series with $q=(x-1)^2$

If $s:=r^2$ then:$$1+r^2+r^4+\cdots=1+s+s^2+\cdots$$

Find the RHS and substitute.

Let the infinite geometric sequence be:

$$a, ar, ar^2, ar^3,...$$

Then, its even elements are:

$$a, ar^2, ar^4,...$$

which form another GP. Can you now proceed?

• Depending on whether your counting is 0-based or 1-based, your even elements might also be $ar, ar^3, ar^5, ...$, which is also another GP either way. Oct 20, 2017 at 8:55