Lower Bound of Pell Number

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Theorem

For all $n \in \N$:

$P_n \ge 2 \paren {1 + \sqrt 2}^{n - 2}$

where:

$P_n$ is the $n$th Pell number


Proof

The proof proceeds by induction.

For all $n \in \N$, let $\map P n$ be the proposition:

$P_n \ge 2 \paren {1 + \sqrt 2}^{n - 2}$


Basis for the Induction

$\map P 1$ is true, as this just says:

$P_1 = 1 = 2 \paren 2^{-1} \ge 2 \paren {1 + \sqrt 2}^{-1}$

It is also necessary to demonstrate $\map P 2$ is true:

$P_2 = 2 \ge 2 \paren {1 + \sqrt 2}^0$

This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P j$ is true for all $1 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$P_j \ge 2 \paren {1 + \sqrt 2}^{j - 2}$ for $1 \le j \le k$


from which it is to be shown that:

$P_{k + 1} \ge 2 \paren {1 + \sqrt 2}^{k - 1}$


Induction Step

This is the induction step:

As we have already shown $\map P 1$ and $\map P 2$, we just need to prove the result for $k \ge 2$.

\(\ds P_{k + 1}\) \(=\) \(\ds P_{k - 1} + 2 P_k\) Definition of Pell Numbers
\(\ds \) \(\ge\) \(\ds 2 \paren {1 + \sqrt 2}^{k - 3} + 4 \paren {1 + \sqrt 2}^{k - 2}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds 2 \paren {1 + \sqrt 2}^{k - 3} \paren {1 + 2 \paren {1 + \sqrt 2} }\)
\(\ds \) \(=\) \(\ds 2 \paren {1 + \sqrt 2}^{k - 3} \paren {1 + 2 \sqrt 2 + \paren {\sqrt 2}^2}\)
\(\ds \) \(=\) \(\ds 2 \paren {1 + \sqrt 2}^{k - 3} \paren {1 + \sqrt 2}^2\) Square of Sum
\(\ds \) \(=\) \(\ds 2 \paren {1 + \sqrt 2}^{k - 1}\)

So $\map P 1 \land \map P 2 \land \dots \land \map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.


Therefore:

$\forall n \in \N: P_n \ge 2 \paren {1 + \sqrt 2}^{n - 2}$

$\blacksquare$

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