# Convergence of a related sequence

Let $(u_n)$ be a sequence such that $u_n^2 = 1$ for all naturals $n$.

Define a sequence $(v_n)$ by $v_n := \displaystyle\sum_{k=0}^n \dfrac{u_0\times u_1\times \cdots \times u_k}{2^k}$.

It is asked to prove that $(v_n)$ converges and that the square of its limit equals $4$.

For the first part, It's not hard to prove that $(v_n)$ is Cauchy thus it converges.

Any ideas, hints for the second part ?

thanks.

The statement is not true. $(v_n)$ is convergent, but the limit can be any value between $-2$ and $2$. Consequently, $\lim_{n \to \infty} v_n^2$ can by any value between $0$ and $4$.
For a proof, let $x \in [-2, 2]$ and choose a base-2 representation of $1 + \frac x2 \in [0, 2]$: $$1 + \frac x2 = p_0 . p_1 p_2 p_3 \ldots = \sum_{k=0}^\infty \frac{p_k}{2^k} \$$ with $p_k \in\{0, 1\}$. Then $$x = -2 + 2 \sum_{k=0}^\infty \frac{p_k}{2^k} = \sum_{k=0}^\infty \frac{2p_k - 1}{2^k} = \sum_{k=0}^\infty \frac{a_k}{2^k}$$ with $a_k = 2p_k - 1 \in \{ -1, 1 \}$. Finally, set $$u_0 = a_0, \\ u_k = \frac{a_k}{a_{k-1}} \text{ for }k > 0\, .$$ so that $a_k = u_0 \times u_1 \times \cdots \times u_k$.