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3. A non-zero converging infinite sum has terms converging to 1

A non-zero converging infinite sum has terms converging to 1

The infinite sum

\begin{equation} p = \prod_{i = 0}^\infty a_i \end{equation}

is defined using the sequence $\{ p_n \}$, with

\begin{equation} p_n = \prod_{i = 0}^n a_i \end{equation}

such that

\begin{equation} p = \lim_{n \to \infty } p_n \end{equation}

As we're working in complete metric spaces ($\mathbb{R}$ or $\mathbb{C}$), if we want our product to converge, we want it to be a Cauchy sequence, so that, for every $\varepsilon > 0$, there exists some integer $N$ such that, for every $m, n > N$,

\begin{equation} |p_m - p_n| < \varepsilon \end{equation}

In particular, for $m = n+1$,

\begin{eqnarray} |p_{n+1} - p_n| &=& (\prod_{i = 0}^{n+1} a_i) - (\prod_{i = 0}^n a_i) \\ &=& (\prod_{i = 0}^n a_i) \left[ a_{n+1} - 1 \right] &<& \varepsilon \end{eqnarray}

There are only two possible cases here : either $p_n$ will be arbitrarily small (but we have assumed that the sum does not converge to zero here), or $a_{n+1} - 1$ does, meaning that $a_{n}$ converges to $1$.