Types of Convergence (Random Variables) 
Suppose that for every $n\ge 1$, the law of $X_n$ is given by $P[X_n=n^2]=\beta_n$ and $P[X_n=0]=1-\beta_n$, determine if $(X_n)_{n\ge 1}$ converges in probability, in $L^1$ or almost sure to zero, when $\beta_n\in\{1/n,1/n^2\}$

Since $P[|X_n-0|>\epsilon]=\beta_n$ $\forall\epsilon>0$, So it converges in probability in both cases.
$\displaystyle E[X_n]=\sum\limits_{\text{only if $k=n^2$}}kP[X_n=k]=n^2\beta_n$, So it converges only if $\beta_n=1/n^2$
Now to almost sure convergence
$P(\lim\limits_{n\to\infty} X_n=0)\overset!=1$
I have a feeling, that this is not just $\lim\limits_{n\to \infty}1-\beta_n=1$, but I don't know why, can you help ?
 A: It depends. The point is that in order to decide whether a sequence of random variables converges almost surely, it does in general not suffices to know the distributions $X_n$ for any $n \in \mathbb{N}$; instead we need to know the joint distributions.
Let's consider the case $\beta_n = \frac{1}{n}$. We define random variables
$$X_n(\omega) := \begin{cases} 0 & x \in [0,1-1/n] \\ n^2 & x \in [1-1/n,1] \end{cases}$$
on the probability space $([0,1],\mathcal{B}([0,1]))$ endowed with the Lebesgue measure. Obviously, $X_n \sim (1-\beta_n) \delta_0 + \beta_n \delta_{n^2}$. Moreover, for any $\omega \in [0,1)$ we can find $n_0 \in \mathbb{N}$ such that $X_n(\omega)=0$ for all $n \geq n_0$. This means that $X_n \to 0$ almost surely.
Now consider
$$Y_n(\omega) := \begin{cases} n^2 & x \in I_n \\ 0 & x \notin I_n \end{cases}$$
where $I_n$ is chosen such that $|I_n|= 1/n$ and for any $\omega \in [0,1]$ and $N \in \mathbb{N}$ we can find $n \geq N$ such that $\omega \in I_n$. As $\sum_n 1/n=\infty$, we may choose $$ \begin{align*} I_1 &:= [0,1],\\ I_2 &:= \left[0,\frac{1}{2} \right], \\ I_3 &:= \left[\frac{1}{2}, \frac{1}{2}+\frac{1}{3} \right] = \left[\frac{1}{2},\frac{5}{6} \right], \\ I_4 &= \left[\frac{5}{6},1 \right] \cup \left[0,\frac{1}{4}-\frac{1}{6} \right]=\left[\frac{5}{6},1\right] \cup \left[0,\frac{1}{12} \right]\\ \vdots & \end{align*}$$ By definition, $X_n \sim Y_n$, but $(Y_n)_n$ does not converge almost surely to $0$.
