Limit of probability of uniform distribution

Question

Let $U_1,U_2,\ldots,U_n$ be independent and identically distributed random variables each following a uniform distribution $U(0, 1).$ How do I find $\lim_{n\to\infty} P(U_1+U_2+\cdots+U_n\le 3n/4)$?

Answer 1

\begin{align} & \operatorname E(U_1+\cdots+U_n) = \frac n2. \\[8pt] & \operatorname{sd}(U_1+\cdots+U_n) = \text{constant} \times \sqrt n. \\ & \quad\text{(where “constant” means not changing as $n$ changes).} \end{align} Ask yourself how many standard deviations above the mean $3n/4$ is, and what Chebyshev's inequality says about the probability of being that many standard deviations above the mean.

Answer 2

Hint: This is a consequence of the law of large numbers after a suitable rewriting of your expression.