You play a sequence of $s$ games, where $s ≥ 2$ is fixed. The outcomes of the various games are independent of each other. The probability that you will win the $k$th game is $\frac{1}{k}$ for $k = 1, 2, . . . , s$. You get one dollar each time you win two games in a row. What is the expected value of the total amount you will get?
I can write the total profit like $\sum_{k=2}^{s}A_k$, with $A_k$ the profit in the $k$th game. I have to find the expected value of the total profit, so $\mathbb{E}[\sum_{k=2}^{s}A_k]=\sum_{k=2}^{s}\mathbb{E}[A_k]$.
Let $V\perp F$ be the events that I win and I loss. Since $\mathbb{P}(V_1)=1$:
$\mathbb{E}[A_1=1]=\mathbb{P}(A_1=1)=\mathbb{P}((V_1)V) =\mathbb{P}(V)=\frac{1}{2},k=1$
$\mathbb{E}[A_2=1]=\mathbb{P}(A_2=1)=\mathbb{P}((V_1)VV) =\mathbb{P}(VV)=\frac{1}{2}\frac{1}{3},k=2$.
So $\mathbb{E}[A_k=1]=\frac{1}{k(k+1)},k\in [1;2]$. Now I can imagine $\mathbb{E}[A_3=2]=\operatorname{1$}+\mathbb{E}[A_3=1]$, so $\mathbb{E}[A_k]=\frac{1}{k(k+1)}$.
The problem is: how do I calculate $\sum_{k=2}^{s}\frac{1}{k(k+1)}$?
Thanks in advance.
Well just telescope: $$ \sum_{k = 2}^s \frac{1}{k(k + 1)} = \sum_{k = 2}^s \frac{(k+1)-(k)}{k(k + 1)} = \sum_{k = 2}^s \left( \frac{1}{k} - \frac{1}{k + 1} \right) = \frac{1}{2} - \frac{1}{s + 1} $$