<s>[INST] proper formula for this geometric series I'm trying to properly write the expression for this geometric series that converges to $\frac{1}{2}$ for all x > 1. It looks like this:
$x = 2, f(x) = \frac{1}{2}$
$x = 3, f(x) = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}$
$x = 4, f(x) = \frac{1}{24} + \frac{2}{24} + \frac{3}{24} + \frac{6}{24} = \frac{1}{2}$
etc.
What is the general equation for this? I can see that the number of actual terms is $1$ term for $x=2$, $2$ terms for $x=3$, $4$ terms for $x=4$, so I figured that the number of terms was $2^{x-2}$. To sum the fractions, the denominator of each term can be $x!$ and the numerator follows the sequence $\{1,2,3,6...\}$ which I can't identify, which is where I'm stuck.
So something like this is... or am I on the wrong track entirely?
$\displaystyle\sum^{2^{x-2}}_{k=1} = \frac{1,2,3,6...}{x!} = \frac{1}{2}$
Any help appreciated. Thanks!
 [/INST] From a "simple solution that meets the requirements" approach, I would recommend using numerators $x-1$ multiplied by each of the numerators from the sum for $f(x-1)$.  So for $x=5$ these additional numerators would be $4,8,12,24$ for the sum as $\frac{1+2+3+6+4+8+12+24}{120}$.  Breaking down individual terms could be somewhat tedious in terms of finding a non-recursive term definition.  From a value perspective, the recursive presentation might look like
$$f(x)=\frac{f(x-1)}x+\frac{(x-1)f(x-1)}x$$
which is obviously of the form $f(x)=f(x-1)$.
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