The Limit of a Recursive Sequence from a Probability Question If we define a sequence as $e_0=0$, for any $n\geq 1$ we have 
$$e_n=\frac{1}{2^n}\left(\sum_{i=0}^{n}C_n^i (e_i +1)\right)$$
How to calculate the limit 
$$\lim_{n\rightarrow\infty}\frac{e_n}{\log_2 n}$$
which is conjectured to be $1$ from a probability question. 
 A: Yes, the limit is indeed equal to $\color{red}{1}$. [I'm writing $\binom{n}{i}$ in place of $C_n^i$ everywhere below.]
The recurrence relation (note that it contains $e_n$ in the RHS too!) can be rewritten as $$e_n=1+2^{-n}\sum_{k=1}^{n}\binom{n}{k}e_k.$$ Introducing $f(x)=\sum\limits_{n=1}^\infty e_n\frac{x^n}{n!}$ and substituting the expression for $e_n$ above, we find $$f(x)=e^x-1+\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{e_k(x/2)^n}{k!(n-k)!}=e^x-1+e^{x/2}f(x/2).$$
Then, $g(x)=e^{-x}f(x)$ satisfies $g(x)=1-e^{-x}+g(x/2)$, which gives $$g(x)=\sum_{k=0}^\infty(1-e^{-x/2^k})=\sum_{k=0}^\infty\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{2^{kn}n!}=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n!(1-2^{-n})}.$$ Plugging this back into $f(x)=e^x g(x)$ (as a product of power series) gives $$\color{blue}{e_n=\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k-1}}{1-2^{-k}}}.$$
Now we have a recursion-free representation, with the Nørlund–Rice method applicable: $$e_n=-\frac{n!}{2\pi i}\oint_\lambda\frac{dz}{z(1-z)\ldots(n-z)(1-2^{-z})}$$ with a simple contour $\lambda$ encircling $z=1,\ldots,z=n$ (and no other poles of the integrand). If we replace $\lambda$ by the rectangular contour with vertices at $z=(2N+1)(\pm 1\pm\pi i/\ln 2)$, where $N$ is a large positive integer, the integral tends to zero when $N\to\infty$. Thus, $e_n$ is equal to the sum (over $m\in\mathbb{Z}$) of residues, at $z=\frac{2m\pi i}{\ln 2}$, of $$\left(z(1-2^{-z})\prod_{k=1}^{n}(1-z/k)\right)^{-1}.$$
Computing these residues, we finally obtain $$\color{blue}{e_n=\frac{1}{2}+\frac{1}{\ln 2}\sum_{k=1}^{n}\frac{1}{k}+\sum_{m\neq 0}\frac{1}{2m\pi i}\prod_{k=1}^{n}\left(1-\frac{2m\pi i}{k\ln 2}\right)^{-1}}.$$
The required asymptotics $e_n\asymp\log_2 n$ is coming from the second term: $\sum\limits_{k=1}^{n}\dfrac1k\asymp\ln n+\gamma+\ldots$
As for the last sum, observe that $$\lim_{n\to\infty}\exp\Big(-\frac{2m\pi i}{\ln 2}\ln n\Big)\prod_{k=1}^{n}\Big(1-\frac{2m\pi i}{k\ln 2}\Big)^{-1}=\Gamma\Big(1-\frac{2m\pi i}{\ln 2}\Big);$$ thus the absolute value of each term of that sum has a finite $n\to\infty$ limit (expressed in terms of $\sinh$, and decaying exponentially with $m\to\infty$), but the term itself is oscillating (slower and slower as $n$ grows).
