Series of equally distributed random variables converges in probability only if each of them is $0$ almost surely Let ${\left( {{X_n}} \right)_{n \in \mathbb{N}}}$ be a sequence of identically distributed random variables such that the series $\sum\limits_{i = 1}^\infty  {{X_i}} $ converges in probability. Show that ${X_n} = 0$ almost surely for all $n \in \mathbb{N}$.
My attempt: The sequence $\sum\limits_{i = 1}^n {{X_i}} $ converges in probability to $X$ if for all $\varepsilon  > 0$, $\mathop {\lim }\limits_{n \to \infty } \mathbb{P}\left( {\left| {\sum\limits_{i = 1}^n {{X_i}}  - X} \right| \geqslant \varepsilon } \right) = 0$.
We know that there exists a sub-sequence ${\left( {\sum\limits_{i = 1}^{{n_k}} {{X_i}} } \right)_{k \in \mathbb{N}}}$ such that $\mathop {\lim }\limits_{k \to \infty } \mathbb{P}\left( {\sum\limits_{i = 1}^{{n_k}} {{X_i}}  = X} \right) = 1$. 
We also know that $\mathop {\lim }\limits_{n \to \infty } {F_{\sum\limits_{i = 1}^{{n_k}} {{X_i}} }}\left( x \right) = {F_X}\left( x \right),\forall x \in C\left( {{F_X}} \right)$, where $C\left( {{F_X}} \right)$ is a set of points of continuity for ${{F_X}}$.
Since $X$ might be a generalized random variable, I'm not sure if I can assume that $\sum\limits_{i = 1}^n {{X_i}\left( \omega  \right)} $ converges (as a sequence of real numbers) almost surely.
I think that just a hint in the right direction should be enough for me to show this. Using the central limit theorem in any of its forms is not allowed.

Hints: Set $S_n := \sum_{i=1}^n X_i$.


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*Fix $\epsilon>0$. It follows from the convergence in probability that $$\lim_{n \to \infty} \sup_{m \geq n} \mathbb{P}(|S_n-S_m|>\epsilon) = 0.$$

*Deduce from the first step and $$\mathbb{P}(|X_1|>\epsilon) = \mathbb{P}(|S_{n+1}-S_n|>\epsilon)$$ that $$\mathbb{P}(|X_1|>\epsilon)=0.$$

*Conclude.

