Another proof by mathematical induction

If $a_1, a_2, a_3, \ldots$ is a sequence of real numbers with $a_i > 0$ for all $i$ , and if $c$ is a fixed real number with $c > 0$ , such that

$a_n \leq ca_{n-1} \qquad \text{for all } n \geq 2$

prove by induction that

$a_n \leq a_1 c^{n-1} \qquad \text{for all } n \geq 1.$

Proof. First, for the case $n=1$ , we have $a_1$ on the left, and on the right we have $a_1 c^0 = a_1$ . Hence, the inequality (which is not strict) holds for this case.
Assume then that the inequality holds for some integer $k \in \mathbb{Z}_{>0}$ . Then, we have

$\begin{align*} &&a_k &\leq a_1 c^{k-1}\\ \implies && ca_k &\leq a_1 c^k & (\text{Multiplying by } c)\\ \implies && a_{k+1} & \leq a_1 c^k & (a_{k+1} \leq ca_k \text{ by assumption}). \end{align*}$

Thus, the statement holds for all $n. \qquad \blacksquare$

Stumbling Robot

Another proof by mathematical induction

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