How to find the limit of a recursive sequence involving multiple recursions? Given $f(1)=1$, what is the limit of the following recursive function as $x$ approaches $\infty$?
$$f(x)=\left(\sum_{n=0}^\infty {1\over{2^{xn}}}\right)\left(\sum_{i=1}^{x-1} {(^x_i)\over2^x}\cdot f(i)\right) $$
where $(^x_i)$ is $xCi$. I've graphed the first 300 or so values of the function and know it's oscillating (like damped harmonic progression) towards a certain number around $0.7213$ish, so I'm fairly certain it's convergent (although I don't know how to prove that on paper).
Also, apologies for the confusing parentheses, I wasn't sure if it was proper to use brackets or something.
 A: [Partial answer for now]
I'll write $f(n)$ instead of $f(x)$, for the sake of clarity, although $f_n$ might be a better notational choice. Anyway...
The recurrence in OP can be simplified, after summing the infinite geometric series in the first multiplicand, to the following recurrence:
$$\forall n\ge 2\colon\quad(2^n-1)f(n)=\sum_{i=1}^{n-1}\binom ni f(i)\tag 1$$
Replacing $n$ by $n+1$ in $(1)$, we also get,
$$\forall n\ge 1\colon\quad(2^{n+1}-1)f(n+1)=\sum_{i=1}^n\binom{n+1}i f(i)\tag 2$$
Subtracting $(1)$ from $(2)$ and using Pascal's rule, we get,
$$\forall n\ge 2\colon\quad(2^{n+1}-1)f(n+1)-(2^n-1)f(n)=f(1)+(n+1)f(n)+\sum_{i=1}^{n-2}\binom ni f(i)\tag3$$
Using $(1)$, we can rewrite $(3)$ as,
$$\forall n\ge 2\colon\quad (2^{n+1}-1)\bigl[f(n+1)-f(n)\bigr]=f(1)+n\bigl[f(n)-f(n-1)\bigr]\tag4$$
Define $\Phi(n):=f(n)-f(n-1)$ for $n\ge 2$ and with $f(1)=1$, we can rewrite $(4)$ as,
$$\forall n\ge 2\colon\quad(2^{n+1}-1)\Phi(n+1)=1+n\Phi(n)\qquad,\quad\Phi(2)=f(2)-f(1)=-\frac 13\tag5$$
and this might be an easier recurrence relation to tackle with, and then get a closed form for $f(n)$. We may even, perhaps, be able to use the telescopic properties of $\Phi(\cdot)$, ie, $\sum_{k=2}^n\Phi(k)=f(n)-f(1)$ with the recurrence in $(5)$, to recover a closed form for $f(n)$. For now, I'm exhausted after an all-nighter and I'll look into it further later.
