Let $a\_1,a\_2,\dots$ be a sequence of positive numbers less than $1$, such that $$\sum\_{n=1}^\infty a\_i= \infty,$$ and $S^1 = \mathbb{R}/\mathbb{Z}$.


Suppose $I\_1,I\_2,\dots$ be random intervals with respective lengths $a\_1,a\_2, \dots$in $S^1$ such that the distribution of the centers of $I\_n$ (for every $n$) are uniform and independent.


It can be shown that with probability $1$, $I = \cup\_{n=1}^\infty I\_n$ is a full measure subset of $S^1$. Is it true that "With probability $1$, $I = S^1$"? If this is not always true, does there exist a good characterization of the sequences $\{ a\_n\}\_{n=1}^{\infty}$ with this property? 


**Edit**. A more precise question: "What happens in the special case $a\_n = \frac1n$?" 



Union of random intervals with total length equal to infinity
This is a refinement of Iosif Pinelis's answer, so we shall be somewhat brief. For a punchline, jump to the "Added" section below.


We claim that if $a\_n>c/n$ holds some $c>1$ and for all $n\geq n\_0$, then $P(I=S^1)=1$. To see this, fix a large integer $N$ and any interval $J$ of length $1/N$. Then,
$$P(J\not\subseteq I)\leq\prod\_{n\_0\leq n<cN}\left(1-\left(\frac{c}{n}-\frac{1}{N}\right)\right)<\exp\left(\sum\_{n\_0\leq n<cN}\left(\frac{1}{N}-\frac{c}{n}\right)\right)=O\_c(N^{-c}).$$
As $S^1$ is a union of $N$ intervals of length $1/N$,
$$P(I\neq S^1)=O\_c(N^{1-c}).$$
The right hand side tends to zero as $N\to\infty$, hence $P(I\neq S^1)=0$ as claimed.


**Added.** Using Google I found out that Shepp (1971) aswered the question completely (see [here](https://link.springer.com/article/10.1007/BF02789327)). In particular, $P(I=S^1)=1$ holds when $a\_n\geq 1/n$ for all $n\geq n\_0$.


