convergence of random variables This is related to this question. How exactly does one go about computing the limits in the answer to the linked question. 
Thanks. 
P.S: I would have commented on the linked question, but I don't have enough points.

So we  have the sequence of constant variables $X_n = 1+1/n$ ($X_n (\omega) = 1+1/n$ for any $\omega \in \Omega$). 
how does one go about showing the following:
 a)
$$ 
\mathop {\lim }\limits_{n \to \infty } P(1 + 1/n \le x) = P(1 \le x),
$$ 
for any $x \neq 1$, showing that the sequence converges in distribution.
 b)
$$
\mathop {\lim }\limits_{n \to \infty } P(|(1 + 1/n) - 1| > \varepsilon ) = 0,
$$ 
for any $\varepsilon > 0$. This shows that the sequence converges in probability.
 c) 
$$
P(\lim _{n \to \infty } (1 + 1/n) = 1) = 1.
$$ which shows that the sequences converges almost surely.
 d)
$$
\mathop {\lim }\limits_{n \to \infty } {\rm E}|(1 + 1/n) - 1|^p  = 0.
$$ showing that the sequence converges in the $p$-th moment.
 A: For a), fix any $x \neq 1$. If $x < 1$, then obviously
$$
\mathop {\lim }\limits_{n \to \infty } P(1 + 1/n \le x) = \mathop {\lim }\limits_{n \to \infty } 0 = 0 = P(1 \leq x).
$$
If, on the other hand, $x > 1$, then there exists $N \in \mathbb{N}$ such that $1+1/n < x$ for all $n > N$. Hence, $P(1 + 1/n \le x) = 1$ for all $n > N$, and so obviously
$$
\mathop {\lim }\limits_{n \to \infty } P(1 + 1/n \le x) = 1 = P(1 \leq x).
$$
For b), first fix $\varepsilon > 0$. Note that
$$
\mathop {\lim }\limits_{n \to \infty } P(|(1 + 1/n) - 1| > \varepsilon ) = \mathop {\lim }\limits_{n \to \infty } P(1/n > \varepsilon ).
$$
Since there exists $N \in \mathbb{N}$ such that $1/n < \varepsilon$ for all $n > N$, it holds $P(1/n > \varepsilon )=0$ for all $n > N$, and hence obviously
$$
\mathop {\lim }\limits_{n \to \infty } P(1/n > \varepsilon ) = 0.
$$
For c), 
$$
P(\mathop {\lim }\limits_{n \to \infty } (1 + 1/n) = 1) = P(1 = 1) = 1.
$$
For d),
$$
\mathop {\lim }\limits_{n \to \infty } {\rm E}|(1 + 1/n) - 1|^p  = \mathop {\lim }\limits_{n \to \infty } {\rm E}|1/n|^p  = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n^p }} = 0.
$$
A: Let $\varepsilon >0$ and choose $n$ large enough so that $(1+1/n)-1<\varepsilon$.  Then,
$$
\left| \Pr [1\leq x]-\Pr [1+1/n\leq x]\right| \leq \Pr [1+1/n-\varepsilon <x]-\Pr [1+1/n\leq x]=\Pr [x<1+1/n<x+\varepsilon ]<\varepsilon .
$$
For the second one, you can use the fact that pointwise (almost everywhere) convergence implies convergence in probability.
The third one is the easiest.  It just follows from the fact that $\left\{ \omega \in \Omega |\, \lim _{n\to \infty}[1+1/n]=1\right\} =\Omega$.
The fourth one is pretty easy too.  Just write down the integral:
$$
\mathrm{EV}\left| (1+1/n)-1\right| ^p=\int _\Omega 1/n^pd\Pr =1/n^p.
$$
Hope that helps!
