$X_1, X_2, \ldots$ 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.
\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).
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 $n\to\infty$.
