Prove the limit property of a 'random' walk/gambling problem Suppose a person plays a sequence of independent games. At the $n$th game, he plays with equally with $n$ other people, gaining $n$ units of money with probability $\frac{1}{n+1}$, losing $1$ unit of money otherwise.
Now we can formulate it as follows. Suppose $(X_n)_{n\geq1}$ are independent random variables with $\mathbb{P}(X_n=n)=\frac{1}{n+1}$ and $\mathbb{P}(X_n=-1)=\frac{n}{n+1}$. Obviously it's a fair game. Note total gaining as $G_n=\sum_{i=1}^n X_i$, prove that $\liminf_n G_n=-\infty$ and $\limsup_n G_n=+\infty$. 
I can prove the second one with the help of Borel-Cantelli Lemma, by showing that $\{X_n=n\}$ happens infinitely many times, but I feel stuck when dealing with the first one. Can anyone help? No matter by what methods. I hope that it can be generalized in some sense.
I don't know if we can use knowledges of martingale since we are trying to prove some 'bad' property (divergence) of $G_n$.
 A: I think martingales can help.
Let $a, b$ be such that $a < 0 < b$. Define $T_a = \inf \{n : G_n = a\}$ and $T_b = \inf \{n : G_n \geq b\}$ (we need $\geq$ here because the process may jump over $b$). Let $T = \min\{T_a, T_b\}$. 
Claim: $T \leq 2(b-a)$ with probability 1. Proof: If $T > b-a$, then $a < G_{b-a} < b$. Consider the next $b-a$ steps; either the player loses all the next $b-a$ games and their winnings fall below $a$, or they win at least once and $G_n$ will exceed $b$ if it has not already fallen below $a$ (because their winnings at this point are at least $b-a$).
We note that $G_T = a \cdot 1_{T_a < T_b} + G_{T_b} \cdot 1_{T_b < T_a}$, and clearly $G_{T_b} \geq b$. Since $T$ is uniformly bounded, we can apply the optional stopping theorem to see that $\mathbb E[G_T] = 0$. Hence:
$$ 0 = \mathbb E[G_T] 
 \geq a \cdot \mathbb P(T_a < T_b) + b \cdot \mathbb P(T_b < T_a) $$
and since $\mathbb P(T_b < T_a) = 1 - \mathbb P(T_a < T_b)$, this shows that
\begin{align*}
   0 &\geq a \mathbb P(T_a < T_b) + b(1 - \mathbb P(T_a < T_b)) \\
\implies \mathbb P(T_a < T_b) &\geq \frac{b}{b-a}.
\end{align*}
From here; fix an arbitrary $a$, and use your previous observation that $\limsup G_n = \infty$ to take a limit of the above estimate as $b \to \infty$ and argue that $T_a < \infty$ with probability $1$.
