Let $X_n$ a sequence of random variables such that $X_n\sim \text{Poisson}(\frac 1n)$. Study the almost-sure convergence of $X_n$.
Since $X_n$ is integer-valued and $P(X_n=0) = \exp(-\frac 1n)$ it is easy to prove that $X_n$ converges to $0$ in probability.
Note that $P(X_n\geq 1) = 1-\exp(-\frac 1n)\sim \frac 1n$, hence $\sum_n P(X_n \geq 1) = \infty$. If the $X_n$ are independent, Borel-Cantelli lemma yields $P\left(\limsup_n \left(X_n\geq 1\right)\right)=1$, hence $X_n$ does not converge to $0$ almost surely.
What can be said when the $X_n$ are not independent ?
If the events $(X_n\geq 1)$ are negatively correlated, a stronger version of Borel-Cantelli (derived from Kochen-Stone lemma) still yields $P\left(\limsup_n \left(X_n\geq 1\right)\right)=1$ (see this).
If $X_n\to 0$ a.s, then $P(\liminf_n (X_n=0))=1$ but I haven't been able to get anything useful out of this.
Here is an example where the almost sure convergence to zero holds. Let $\left(\xi_i\right)_{i\geqslant 1}$ be an independent sequence of random variables, where $\xi_i$ has a Poisson distribution of parameter $i^{-1}-(i+1)^{-1}$. Since the $\xi_i$ are $\geq 0$ a.s., the series $\sum_{n\geq 1} \xi_n$ converges a.s. in $\mathbb R\cup\{\infty\}$. Besides, $$E\left(\sum_{n\geq 1} \xi_n\right) = \sum_{n\geq 1} \frac{1}{n(n+1)} <\infty,$$ hence $\sum_{n\geq 1} \xi_n$ converges a.s. in $\mathbb R$. It is therefore licit to define $X_n:=\sum_{i\geqslant n}\xi_i$, and by what precedes $X_n$ converges almost surely to $0$.
It remains to prove that $X_n\sim \mathcal P(\frac 1n)$. Let $n\geq 1$ and $s\geq 0$ be fixed. Let $A_m$ denote the event $$A_m=\left\{\sum_{i=n}^m \xi_i=s \right\}\cap \bigcap_{p\geq m}\{\xi_p=0\},$$
so that $A_m\subset A_{m+1}$, and $P(X_n=s) = P(\bigcup_{m\geq n}A_m) = \lim_m P(A_m).$ By independence of the $\xi_i$ and a classical result for finite sums of independent Poisson r.v., $$\begin{aligned} P(A_m) &= P(\sum_{i=n}^m \xi_i=s) P(\bigcap_{p\geq m}\{\xi_p=0\}) \\&= \frac{1}{s!}\left(\frac 1n - \frac 1{m+1}\right)^s \exp\left(-(\frac 1n - \frac 1{m+1})\right) P(\bigcap_{p\geq m}\{\xi_p=0\}). \quad \quad \quad \quad (1) \end{aligned}$$
Note the estimate $$\begin{aligned} 1-P(\bigcap_{p\geq m}\{\xi_p=0\}) &\leq \sum_{p\geq m} P(\{\xi_p\neq 0\}) \\&= \sum_{p\geq m} 1-\exp(\frac 1p - \frac 1{p+1}) \\&\sim \sum_{p\geq m}\frac 1{p^2} \\&=o_m(1). \end{aligned} $$
Returning to $(1)$, we obtain that $$P(X_n=s) =\lim_m P(A_m) = \frac{1}{s!}\frac 1{n^s} \exp(-\frac 1n ),$$ thus $X_n\sim \mathcal P(\frac 1n)$.