$R_{n} = \lVert X_{n} \rVert \to 1$ in probability, where $ X_{n}$ has uniform distribution in the $n$-dimensional ball. Using the transformation method, with the polar coordinates transformation, it is
easy to show that the random variable $R_{2} = \lVert X_{2}\rVert$ has cumulative distribution
$$
F_{2}(x)= P(R_{2} \leq 2)
=
\begin{cases}
 0, & x<0,\\
x^{2}, & 0\leq x \leq 1\\
1, & x>1.
\end{cases}
$$
In the same way, using spherical coordinates we have that the cumulative distribution of $R_{3}=
\lVert X_{3} \rVert$ is
$$
F_{3}(x)
=
P(R_{3}\leq x)=
\begin{cases}
0 ,& x < 0,\\
x^{3},& 0\leq x \leq 1,\\
1, & x>1.
\end{cases}
$$
I don't know how to compute the distribution of $R_{n}$ for $n \geq 4$, but I conjecture
that it is
$$
F_{n}(x) = P(R_{n} \leq x)
=
\begin{cases}
0,& x< 1\\
x^{n},& 0 \leq x \leq 1\\
1, & x > 1.
\end{cases}
$$
If it is true, we can find that, for every $\epsilon > 0$
$$
\begin{align*}
P(\lvert R_{n} - 1\rvert \geq \epsilon) &= 1 - P(\lvert R_{n} - 1 \rvert < \epsilon)\\
&=1 - (F_{n}(1+\epsilon) -F_{n}(1-\epsilon))\\
&=
\begin{cases}
0 & \epsilon > 1\\
(1 - \epsilon)^{n} & 0 < \epsilon < 1.
\end{cases}
\end{align*}
$$
Then, $P(\lvert R_{n} - 1\rvert \geq \epsilon) \to 0$ as $n \to \infty$.
So I need to prove if my conjecture is true or false in order to solve the proposed problem.
 A: Let $``\text{vol''}$ denote the $n$-dimensional
Lebesgue measure on $\mathbb{R}^{n}$. Since $X_{n}$ has uniform distribution in
the $n$-dimensional ball, $B_{n}(\mathbf{0},1)$,
we have that
$$
P(X_{n} \in A)=
\int_{A}
\frac{1}{\text{vol}(B_{n}(\mathbf{0}, 1))}
\chi_{B_{n}(\mathbf{0}, 1)}\,
d\text{vol}=
\frac{\text{vol}(B_{n}(\mathbf{0},1)\cap A)}
{\text{vol}(B_{n}(\mathbf{0}, 1))}
$$
for all  measurable set $A \subset \mathbb{R}^{n}$.
Let's compute the volumen of a ball centered at origin and with radius $r$, $B_{n}(\mathbf{0}, r)$,
in terms of the volumen of unitary ball $B_{n}(\mathbf{0}, 1)$. In fact, by the change of
variable $(u_{1}, \dots, u_{n}) = (r x_{1}, \dots, r x_{n})$ the unitary ball $B_{n}(\mathbf{0}, 1)$
is transformed in one-to-one fashion into the ball
$B_{n}(\mathbf{0}, 1)$, in addition, the jacobian
of this transformation is $r^{n}$. Then,
$$
\text{vol}(B_{n}(\mathbf{0}, r))
=
\int_{B_{n}(\mathbf{0},r)}
\, d\text{vol}
=
\int_{B_{n}(\mathbf{0}, 1)}
r^{n}\,
d\text{vol}
=
r^{n}\text{vol}(B_{n}(\mathbf{0},1))
$$
Hence, the distribution of $R_{n}$ is given by
$$
\begin{align*}
F_{R_{n}}(r) &=
P(\lVert X_{n} \rVert \leq r)\\
&= 
P(X_{n} \in B_{n}(\mathbf{0}, r))
&=
\begin{cases}
0, & r \leq 0,\\[1em]
\dfrac{\text{vol}(B_{n}(\mathbf{0}, r))}
{\text{vol}(B_{n}(\mathbf{0}, 1))}=r^{n},
& 0 < r \leq 1,\\[1em]
\dfrac{
\text{vol}[B_{n}(\mathbf{0}, r) \cap B_{n}(\mathbf{0},1)]
}
{\text{vol}(B_{n}(\mathbf{0},1))} = 1, & r > 1.
\end{cases}.
\end{align*}
$$
Finally,
$$
P(\lvert R_{n}- 1\rvert \geq \epsilon)
= 1- P(\lvert R_{n} - 1 \rvert < \epsilon)
= F_{R_{n}}(1 - \epsilon).
$$
From this is easy to see that $P(\lvert R_{n} - 1\rvert \geq \epsilon) \to 0$ as $n \to \infty$.
