Probability

They're not equivalent.

$\displaystyle\sum_{k=1}^\infty \left(\frac{1}{2}\right)^k \left(\frac{11}{12}\right)^{k-1} = \sum_{k=1}^\infty \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)^{k-1} \left(\frac{11}{12}\right)^{k-1} = \sum_{k=1}^\infty \left(\frac{1}{2}\right) \left(\frac{11}{24}\right)^{k-1} = \frac{1}{2} \sum_{k=1}^\infty \left(\frac{11}{24}\right)^{k-1}$

I agree with ttoby. I'd suggest posting a photo of the actual question.

EDIT2: Nevermind got it sorted now, but have another question about this! How do I simplify the summation from k=1 to infinity of (11/14)^k-1?

Here's the question:

I got up to the summation from k=1 to infinity of (1/2)^k (11/12)^(k-1)
EDIT: Got rid of the question, mind if I pm you the question instead? Just found out I can't make the question publicly available...
EDIT2: Nevermind got it sorted now, but have another question about this! How do I simplify the summation from k=1 to infinity of (11/14)^k-1?

Setting aside the factor of 1/2 you have a geometric progression. First term is? Common ratio is? Apply forumla a/(1-r) = result. Then divide by 2.

It's the sum of an infinite geometric series. $1 + (11/14) + (11/14)^2 + ... = \dfrac{1}{1-\frac{11}{14}}$