Given a fair coin is tossed $n$ times and the resulting sequence contains no two consecutive heads, Let $E(n)$ be the expected number of heads, what is $\displaystyle \lim_{n\rightarrow\infty} \frac{E(n)}{n}$
Here's my approach
For $n$ coin tosses, $E(n) = \displaystyle \sum_{k=0}^{n} k \mathrm{Prob}(\mathrm{exactly} \; $k$ \; \mathrm{heads} | \mathrm{sequence \ has \ no \ consecutive\ heads})$
# of sequences of length $n$ with no consecutive heads can be counted easily using fibonacci recursion and it is $F_{n+2}$ (where $F_0=0, F_1=1$)
And for # sequences with exactly k heads s.t. no two heads are consecutive is equal to # of solutions to this equation $x_1 + x_2 + \ldots + x_k + x_{k+1} = n-k$, where $x_1, x_{k+1} \geq 0$ and $x_j \geq 1 \; \forall \; 2 \leq j \leq k$
It has exactly $\binom{n-k+1}{k}$ solutions (bars and stars)
Thus, $E(n) = \displaystyle \frac{1}{F_{n+2}}\displaystyle \sum_{k=0}^n k \binom{n-k+1}{k} $ (I know the latter half is $0$ for $k>n/2$)
I don't know how to proceed further, any help will be appreciated