If $(X_n)$ are i.i.d random variables, then $P(X_1\geq 0)=1$ and $P(X_1 >0)>0$ implies $\sum X_n =\infty$ 
Let $(X_n)$ a sequence of i.i.d random variables, such that $P(X_1 \geq 0)=1$ and $P(X_1 >0) > 0$. Show that $\sum_{n \geq 1}X_n =\infty$ almost surely.

We have that $P(X_1 > 0)= \underset{n \rightarrow \infty}{\lim}P(X_1 > \frac{1}{n})$. So by the property of the limit, there exists a sufficiently big $N \in \mathbb{N}$ such that $\forall n \geq N$ we have $P(X_1 > \frac{1}{n})>0$. Let then put $A_n = \{X_1 \geq \frac{1}{N} \}$. We then have:
$$\sum_{n \geq 1} X_n \geq \sum P(A_n) \geq M + \sum_{n\geq N} P(X> \frac{1}{N}) = M + P(X>\frac{1}{N})\sum_{n\geq N}1$$
Which equals to infinity. Is my reasoning correct?
I am very unsure, as $\sum_{n\geq 1}X_n$ is ambiguous to me, I am unsure what it exactly represents. 
 A: A necessary condition for $\sum_{n \geq 1}X_n$ to converge is to have $X_n \to 0$ a.s.. With your little argument, we know that $\exists \alpha >0 $ s.t. $ P(X_n>\alpha)>0.$. Then (by second Borel-Cantelli lemma) :
$$P(X_n>\alpha \text{ infinitely often})=1.$$
Then $P(\sum_{n \geq 1}X_n \text{ diverges })=1$.
A: The exercise asks you to show that the set $\lbrace \omega \in \Omega \vert \sum_n X(\omega) \text{ diverges} \rbrace $ has measure 1.
Choose N such that $\mathbb{P}(X_1 > 1/N) = \alpha > 0$. 
Tail sigma algebra approach: Now consider the complement case where the sequence converges. This is contained in the event "only finitely many $X_n$'s are greater than $1/n$ which is a tail event. Then it only remains to show that the probability of this event is not 1 and then work out what that means for our original set. 
Alternatively try applying Borel-Cantelli II.
A: $\sum_n X_n$ is another random variable, just as $X_1, X_2$, etc.; but there is a relevant detail though.
Since for every element $\omega$ in the sample space $\Omega$ (that is, for every possible result of the random experiment that undergoes the whole situation and calculations) each $X_n$ takes some real value $X_n(\omega)$ —remember that the $X_n$ random variables are by definition functions
$$X_n\colon \Omega \longrightarrow \mathbb R$$
(for each $n\in\mathbb N$)— then for each $\omega$ also $\sum_n X_n$ takes a value, namely,
$$\sum_n X_n(\omega)=X_1(\omega)+\cdots+X_n(\omega)+\cdots.$$
BUT: it could be the case that this series is divergent, and in that case the value of $\sum_n X_n (\omega)$ would be $\infty$. Anyway, it's just a matter of saying that this is a r.v. which takes values not in $\mathbb R$ but in $$\overline{\mathbb R}=\mathbb R \cup \{\infty\}.\quad(*)$$ 
Now, to say that an event $A$ happens almost surely is just to say that $P(A)=1$. So you are just asked to show that
$$P\left(\sum_n X_n = \infty\right)=1$$
that is
$$P(A)=1,$$
$A$ being
$$A=\{\omega \in \Omega \colon \sum_n X_n(\omega) \text{ diverges}\}.$$
And you can get more and more specific, as in
$$A=\{\omega \in \Omega \colon \forall M>0 \exists N\in \mathbb N(n\ge N \implies X_1(\omega)+\cdots+X_n(\omega) \ge M\},$$
and so on.
This is just theory and an explanation of the concepts involved, but since your hypothesis are not given in terms of 'omegas' and since you actually don't know much about the probability that is given to the sample space, you don't need to make those $\omega$ explicit. You don't even need to write it as probabilities, since you can just work with the $X_n$ and $\sum_n X_n$ as mere numbers as you did, as long as you mention that your conclusion holds 'a.s.' or with probability one.
So in your proof, if you're sure that $\sum_n X_n$ is greater than the last expression almost sure and that this last expression diverges almost sure, then you're done.

(*) The only drawback of this approach is that instead of the borelian sets of $\mathbb R$ we have to consider a $\sigma$-algebra over $\overline{\mathbb R}$ to determine for which sets $B$ the expressión $P(\sum_n X_n \in B)$ makes any sense. But the borelians of $\overline{\mathbb R}$ is a well defined family of sets, so there's no such problem.
If you still wanted to avoid this approach, you would have to write the event $\{\sum_n X_n = \infty\}$ in terms of simpler events which do not involve the idea of taking the value $\infty$.
A: You want to prove that the sum $\sum_n X_n$ diverges with probability $1$. Speaking somewhat informally, it suffices to show that for every integer $k$, $\sum_n X_n > k$ with probability $1$, since the intersection of countably may probability $1$ events has probability $1$. 
I think some of the ideas in your post are already enough to show this weaker statement.
A: There exists $\alpha>0$ so that $p=P(X_1>\alpha)>0$.
Let $E_N$ be the event "$X_n\le\alpha$ for every $n>N$." By independence, $$P(E_N)=\prod_{n>N}P(X_n\le\alpha)=\prod_{n>N}(1-p)=0.$$So $$P(\bigcup_NE_n)=0,$$which says exactly that "$X_n\ge\alpha$ for infinitely many $n$" holds almost surely. Hence $\sum X_n$ diverges almost surely.
