 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Put $u=c^{-1}-1>0$, so $c=(1+u)^{-1}$. By <Bernoulli's inequality>, $(1+u)^n\geq1+nu$, and hence
$$
\boxed{0<c^n\leq\frac1{1+nu}\longrightarrow0.}
$$
More explicitly, for $\varepsilon>0$ choose an <integer> $N>1/(u\varepsilon)$. For $n\geq N$, the right-hand side is less than $\varepsilon$. Thus the <geometric sequence> has <limit of a sequence> zero, either directly by the definition or by the <squeeze theorem>.

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