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Define a sequence as follows: $a_1 = 1$, $a_2 = 3$ and $a_{n+1} = a_{n-1} + a_n$ for integers $ n \ge 2$

Use strong induction to prove that for all integers $n\ge1$, we have $a_n \lt (\frac{7}{4})^n$.

Let $P(n)$ be the proposition $a_n \lt (\frac{7}{4})^n$. I proved $P(1)$ , $P(2)$, $P(3)$, and $P(4)$ as a basis step. How could I start proving $P(k+1)$ assuming that $P(j)$ is true for $1 \le j \le k$ where $k$ is an integer $ k\ge 4$ ? Also I want to know if proving $P(n)$ for $1 \le n \le 4 $ is right as basis step or not ?

Out Of Bounds
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1 Answers1

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We are given the recursion relation \begin{align*} a_1 & = 1\\ a_2 & = 3\\ a_{n + 1} & = a_{n - 1} + a_{n}, n \geq 2 \end{align*} Let $P(n)$ be the statement $a_n < \left(\frac{7}{4}\right)^n$.

We wish to prove $P(n)$ holds for all positive integers $n$. Since each term save the first two is expressed in terms of the preceding two terms, we must prove two base cases. They are $P(1)$ and $P(2)$.

Let $n = 1$. Then $$a_1 = 1 = \frac{4}{4} < \frac{7}{4} = \left(\frac{7}{4}\right)^1$$ so $P(1)$ holds.

Let $n = 2$. Then $$a_2 = 3 = \frac{48}{16} < \frac{49}{16} = \left(\frac{7}{4}\right)^2$$ so $P(2)$ holds.

To prove $P(n)$ by strong induction, we assume that $P(n)$ holds for each integer $n \leq m$, where $m \geq 2$. In particular, we may assume that $P(m - 1)$ and $P(m)$ both hold. With that in mind, we then let $n = m + 1$. \begin{align*} a_{m + 1} & = a_{m - 1} + a_m && \text{by definition}\\ & < \left(\frac{7}{4}\right)^{m - 1} + \left(\frac{7}{4}\right)^m && \text{by the induction hypothesis}\\ & = \left(\frac{7}{4}\right)^{m - 1}\left(1 + \frac{7}{4}\right)\\ & = \left(\frac{7}{4}\right)^{m - 1} \cdot \frac{11}{4}\\ & < \left(\frac{7}{4}\right)^{m - 1} \cdot \frac{49}{16}\\ & = \left(\frac{7}{4}\right)^{m - 1} \cdot \left(\frac{7}{4}\right)^2\\ & = \left(\frac{7}{4}\right)^{m + 1} \end{align*} Since we have established that $P(1)$ and $P(2)$ hold and $P(m + 1)$ holds if $P(n)$ holds for each integer $n \leq m$, $P(n)$ holds for all positive integers.

N. F. Taussig
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