Following game between person A and B is proposed:
Person A has an unfair coin with probability $p \in (\frac{1}{3},\frac{1}{2})$ of heads. Person B starts with a captial of 100€. For each time B tosses heads, his captial doubles and is halved if B tosses tails. Each time A gives or gets the money.
So let $X_n :=$ B's capital after the n'th toss
Now A further says, that $\lim_\limits{n \rightarrow \infty} \mathbb{E}[X_n]=\infty,$
Which is to be proven or disproven.
My Attempt: To check this we define $X_0:=100$ and get the recursion: $$X_n=X_{n-1}(1+W_n), W_n:= \left\{ \begin{array}{ll} 1 & \textrm{B tosses heads}\\ -\frac{1}{2} & \, \textrm{otherwise} \\ \end{array} \right. $$
If we now write out $X_n$, we get $X_n=100 \displaystyle{\prod_\limits{i=1}^n}(1+W_i)$. Now if we define $Q_i:=(1+W_i)$ and note that $Q_1, \dots , Q_n$ are independent random variables, we get that: $$\mathbb{E}[X_n]=\mathbb{E}[100 \displaystyle{\prod_\limits{i=1}^n}Q_i]=100 \displaystyle{\prod_\limits{i=1}^n}\mathbb{E}[Q_i]=100 \displaystyle{\prod_\limits{i=1}^n}(2p+\frac{1}{2}(1-p))=100(\frac{3}{2}p+\frac{1}{2})^n \longrightarrow \infty$$
Question: I think I have done something wrong because, when calculating $\mathbb{E}[X_n]$ without the recursion, so saying
$X_n \sim Bin(n,p,k)$:
$$\mathbb{E}[X_n]=\sum_\limits{k=1}^n k P(X_n=k)=\sum_\limits{k=1}^n100(2)^k(\frac{1}{2})^{n-k}\binom{n}{k}p^k(1-p)^{n-k}=100((\frac{3p+1}{2})^n(\frac{1-p}{2})^n) \longrightarrow \infty $$
Which clearly is not the same expected value I got, with the recursion. I would be very grateful, if anyone could point out my mistake. Thanks in advance.
Edit: As pointed out in the comments of the answer I did not sum over $k=0$, so the correct sum would be: $$\sum_\limits{k=0}^n k P(X_n=k)=\sum_\limits{k=0}^n100(2)^k(\frac{1}{2})^{n-k}\binom{n}{k}p^k(1-p)^{n-k}=100(\frac{3}{2}p+\frac{1}{2})^n$$ which also delivers the same expected value, as in my attempt and the answer.
You did nothing wrong, you can do it with the conditional expectation as well. We have $$E(X_{n+1} | X_n) = p 2 X_n + (1-p) \frac{X_n}{2} = (3/2p - 1/2) X_n$$ so $$E(X_{n+1}) = E(E(X_{n+1} | X_n)) = (3/2p - 1/2) E(X_n)$$ which gives by recurrence $$E(X_n) = 100 (3/2p - 1/2)^n$$.
Also, I helped someone earlier on the same problem on how to prove the almost sure convergence to $0$ of $X_n$ : (The link was deleted, the explanation is in the comments)