Constructing a sequence of random variables that converges in probability but not almost surely in a specific probability space. Problem settings: Given $\Lambda=\left\{ 1,...,k\right\}$
  we mark $\Omega=\Lambda^{\mathbb{N}}$
  and we define a cylinder set to be:$$\left[\lambda_{1},...,\lambda_{m}\right]=\left\{ \omega\in\Omega\;|\;\omega_{i}=\lambda_{i}\:\forall\,1\leq i\leq m\right\}$$ 
 Let $\mathcal{F}$
  be the $\sigma$-
 algebra generated by all cylinders in $\Omega$
  and let $\left(p_{1},...,p_{k}\right)$
  be a probability vector, we define a measure on cylinders by: $$\mathbb{P}\left(\left[\lambda_{1},...,\lambda_{m}\right]\right)=p_{\lambda_{1}}\cdots p_{\lambda_{m}}$$
 We then extend this measure to $\mathcal{F}$
  and receive a probability space $\left(\Omega,\mathcal{F},\mathbb{P}\right)$
 .
The question: I'm looking for a sequence of random variables in this space that converges in probability but not almost-surely. It would suffice if I could define $\left\{ X_{n}\right\} _{n=1}^{\infty}$ to be an independent sequence such that $X_{n}\left(\omega\right)\in\left\{ 0,1\right\}$ 
  for all $\omega\in\Omega$
  and $\mathbb{P}\left(X_{n}=1\right)=\frac{1}{n}$
  for all $n\in\mathbb{N}$
 . This would give me convergence in probability to $0$
  instantly and since $${\displaystyle \sum_{n=1}^{\infty}\mathbb{P}\left(X_{n}=1\right)}={\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}=\infty}$$
  I would get from Borel-Cantelli's lemma that $\mathbb{P}\left(\limsup\left\{ X_{n}=1\right\} \right)=1$
  and thus there is no convergence almost-surely to $0$. My problem is coming up with an actual sequence that would satisfy these conditions. As a side note $\Lambda$ can be taken to be any finite set and $\left(p_{1},...,p_{k}\right)$ can be any probability vector.
 A: I don't see how to define a suitable sequence of random variables satisfying the conditions you mentioned, but here is a different approach:
For $j \geq 1$, we set $$\Lambda^j := \{(\lambda_1,\ldots,\lambda_j,0,\ldots); \forall l=1,\ldots,j: \lambda_l \in \Lambda\}$$
Obviously, $|\Lambda^j|<\infty$. Consequently, $\Lambda^{\infty} := \bigcup_{j \in \mathbb{N}} \Lambda^j$ is countable. We denote by $(\sigma_j)_{j \in \mathbb{N}}$ an arbritrary enumeration. Define
$$X_j(\omega) := \begin{cases} 1 & \forall l \in \mathbb{N}, \sigma_j(l) \neq 0: \omega(l) = \sigma_j(l) \\ 0 & \text{otherwise} \end{cases}$$
Intuitively: For fixed $m \in \mathbb{N}$ there exists only a finite number of possibilities how the first $m$ elements of any element $\omega \in \Omega = \Lambda^{\mathbb{N}}$ might look like. Step by step, we run through all these possibilities.
The construction of the random variables implies in particular that for any $\omega \in \Omega$, $j_0 \in \mathbb{N}$ we can find $j \geq j_0$ such that
$$X_j(\omega)=1$$
Indeed: Fix $\omega \in \Omega$ and define $$\tau_m := (\omega(1),\ldots,\omega(m),0,\ldots)$$ Obviously, $\tau_m \in \Lambda^{\infty}$ for any $m \geq 1$. In particular, there exists $j=j(m)$ such that $\sigma_{j(m)}=\tau_m$. Hence, by definition, $X_{j(m)}(\omega)=1$ for all $m \geq 1$. Since $(j(m))_{m \in \mathbb{N}}$ is unbounded, this proves the claim.
Consequently, $X_j$ does not converge almost surely. 
On the other hand, we have $\left| \bigcup_{m \leq n} \Lambda^m \right|<\infty$ for any $n \in \mathbb{N}$. Therefore, it is not difficult to show that
$$\mathbb{P}(X_j = 1) \leq \max_{1 \leq l \leq k} p_l^{n}$$
for $j \geq j_0(n)$ sufficiently large. Since the right-hand side converges to $0$ as $n \to \infty$, this shows that $\mathbb{P}(X_j = 1) \to 0$ as $j \to \infty$, i.e. $X_j \to 0$ in probability. 
