Counterexample for Converse of Borel-Cantelli Lemma

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The Problem: Let $K$ be a positive integer valued random variable on $(\Omega,\mathcal F,P)$. Define a sequence $\{X_n\}_{n\in\mathbb N}$ of $\{0,1\}$-valued random variables on $(\Omega,\mathcal F,P)$ by $$X_n(\omega)=\begin{cases}0&\text{if }n\leq K(\omega)\\1&\text{if }n>K(\omega).\end{cases}$$ Show that by a suitable choice of the distribution of $K,\,\sum_{n=1}^\infty P(X_n\ne1)=\infty$. In other words, the Borel-Cantelli lemma cannot detect the almost sure convergence $X_n\longrightarrow1.$

My Attempt: Choose a random variable $K$ on $(\Omega,\mathcal F,P)$ with distribution given by $$p_K(k)=\frac{1}{k(k+1)},\quad k\in\{1,2,\dots\}.$$ By construction, $$P(X_n\ne1)=P(n\leq K)=1-\sum_{k=1}^{n-1}\frac{1}{k(k+1)}=1-1+\frac{1}{n}=\frac{1}{n}.$$ Hence, it follows that $$\sum_{n=1}^\infty P(X_n\ne1)=\sum_{n=1}^\infty\frac{1}{n}=\infty.$$ To prove that indeed $X_n\longrightarrow1$ almost surely, note that for a fixed $\omega\in\Omega$, we have $K(\omega)=k$ for some fixed positive integer $k$, hence $X_n(\omega)=1$ for all $n>k$, from which the almost sure convergence follows.


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