Prove that the convergence of $\prod_{n=1}^{\infty} b_n$ implies $\lim_{n→∞} b_n = 1$

Question

So I am stuck on this question and am unsure how to solve it, would be appreciated if someone could help me out, thanks! :)

Question

Consider a sequence $(b_n)_{n=1}^{\infty}$ of non-zero real numbers. By definition, the infinite product $\prod_{n=1}^{\infty}$ converges if the sequence $(p_n)_{n=1}^{\infty}$, where

$p_n=\prod_{k=1}^{n}b_k$

converges to some non-zero number. Prove that the convergence of $\prod_{n=1}^{\infty}b_n$ implies $\lim_{n\to\infty}b_n=1$

Answer 1

Hint:

$$b_n = \frac{\prod_{k=1}^nb_k}{\prod_{k=1}^{n-1}b_k}$$

Answer 2

As comments are suddenly appearing, I thought I'd take a moment to fix the original, flawed answer. Clearly, RRL's answer is perfectly sufficient and quite straight-forward. However, it's sometimes a nice exercise to use the $\epsilon-\delta$ definition for limits, so I'll do that rather than delete the answer altogether.

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Suppose that $p_n \to p$. Then $p_n$ is a Cauchy sequence. Fix $\epsilon' > 0$, and choose $\epsilon > 0$ such that $\epsilon < \min \left\{\frac{|p|}2, K\right\}$, where $K := \min\left\{\left\vert p - \frac{\vert p \vert}2\right\vert, \ \left\vert p + \frac{\vert p \vert}2\right\vert\right\}$. Suppose that $N \in \mathbb{N}$ is such that $n, m \ge \mathbb{N}$ implies $\vert p_n - p_m \vert < \epsilon$. In particular, taking $m \to \infty$, we see that $|p_n - p| < \epsilon$, which can be rewritten as \begin{align*} p - \frac{|p|}2 < p - \epsilon < \ & p_n < p + \epsilon < p + \frac{|p|}2 \\ K < |p_n| < \max&\left\{\left\vert p - \frac{\vert p \vert}2\right\vert, \ \left\vert p + \frac{\vert p \vert}2\right\vert\right\}. \end{align*}

Finally, observing that $p_{n+1} - p_n = p_n(a_{n+1} - 1),$ we see that \begin{align*} |a_{n+1} - 1| = \frac{|p_{n+1} - p_n|}{|p_n|} < \frac{\epsilon}{K} < \epsilon' \end{align*} From which it follows that $a_n \to 1$ as $n \to \infty$.