Independent, random variables with equal distribution satisfy: $\lim_{n \to \infty}\mathbb{P}\left(X_{n+1} > \sum_{i = 1}^{n}X_i\right) = 0$ 
$X_1, X_2, \ldots$ are independent, non-negative, real random variables with equal probability distribution. Show that
  $$\lim_{n \to \infty}\mathbb{P}\left(X_{n+1} > \sum_{i = 1}^{n}X_i\right) = 0.$$

Well I cannot move on with this problem. I would be grateful for any help.
 A: \begin{align}
&P(X_{n+1} >\sum_{i=1}^n X_i)\\
 \leq &P(X_{n+1} > X_i, i = 1,2\cdots,n) \\
\leq &P(X_{n+1} \text{ is the unique maximum among }X_i, i =1 ,2, \cdots, n+1)\\
 \leq &\frac{1}{n+1}
\end{align}
Since the probability that there is a unique maximum is less or equal than 1, and all elements have equal chance to be that(exclusively).
A: If $x_{n+1}>x_1+\cdots+x_n$ and $x_1,\ldots,x_n,x_{n+1}\ge 0$ then $x_{n+1}>x_k$ for every $k\in\{1,\ldots,n\}$.
Above I am not saying anything about random variables but only about non-negative numbers.  Now apply it to random variables:
$$
\Pr\left(X_{n+1} > \sum_{k=1}^n X_k \right) \le \Pr\left( X_{n+1}>X_1\  \&\  \cdots\  \&\  X_{n+1}>X_n \right). \tag 1
$$
If the distribution were continuous, then one could immediately say that that last probability is $1/(n+1)\to0\text{ as }n\to\infty$.  The reason is that the probability that there are two or more of these random variables that are equal to each other would be $0$, so there would always be a unique maximum among them.  The probability that the value of $k$ that is the index of the maximum is equal to any particular one of the numbers $1,\ldots,n,n+1$ is $1/(n+1)$.  Since we don't have the assumption that there are no point masses in this distribution, we go on from where we are above and say that the probability is $(1)$ is
$$
\begin{align}
& \Pr(\text{no ties for first place and the index of the max is }n+1) \\[6pt]
= {} & \Pr(\text{index of max}=n+1 \mid \text{no ties for first place})\cdot\Pr(\text{no ties for first place}) \\[6pt]
\le {} & \frac{1}{n+1}\cdot 1.
\end{align}
$$
Then find the limit of that as $n\to\infty$.
