Monotone Sequences

Reciprocal of the Reciprocal

Let

a, b

be positive real numbers. Set

x_{0} = a

and

x_{n + 1} = {(x_{n}^{- 1} + b)}^{- 1}

for

n \geq 0

Prove that $x_{n}$ is strictly monotone decreasing.
Prove that the limit exists and find it.

For the induction step we'll notice before anything else that in general for any $n \in N_{1}$ we have that $\begin{aligned} x_{n + 1} & = {(x_{n}^{- 1} + b)}^{- 1} \\ = \frac{1}{\frac{1}{x_{n}} + b} \\ = \frac{1}{\frac{1 + b x_{n}}{x_{n}}} \\ = \frac{x_{n}}{1 + b x_{n}} \end{aligned}$ moving on, let $k \in N_{1}$ we'll prove that $x_{k} \geq x_{k + 1}$ , $\begin{aligned} x_{k + 1} & = \frac{x_{k}}{1 + b x_{k}} \\ = \frac{\frac{x_{k - 1}}{1 + b x_{k - 1}}}{1 + b (\frac{x_{k - 1}}{1 + b x_{k - 1}})} \\ = \frac{\frac{x_{k - 1}}{1 + b x_{k - 1}}}{(\frac{1 + b x_{k - 1} + b x_{k - 1}}{1 + b x_{k - 1}})} \\ = \frac{x_{k - 1}}{1 + 2 b x_{k - 1}} \end{aligned}$ But then because of the fact that $1 + 2 b x_{k} > 1 + b x_{k}$ then we know that $x_{k + 1} = \frac{x_{k - 1}}{1 + 2 b x_{k - 1}} < \frac{x_{k - 1}}{1 + b x_{k - 1}} = x_{k}$

Finally as $x_{0} = a > \frac{a}{1 + a b} = x_{1}$ as $a b \in R^{+}$ , then we know that $x_{k} > x_{k + 1}$ for all $k \in N_{0}$

Next we can prove that all terms are positive because $a, b > 0$ and all terms are derived from them using operations that maintain positivity so in the induction step it will work. Therefore it is lower bounded by $0$ by MCT the limit exists.

Recall that $\begin{aligned} lim_{n \to \infty} x_{n} & = L \\ = lim_{n \to \infty} x_{n + 1} \\ = lim_{n \to \infty} \frac{1}{\frac{1}{x_{n}} + b} \\ = \frac{1}{\frac{1}{L} + b} \\ = \frac{L}{1 + L b} \end{aligned}$ We see that this implies $1 + L b = 1$ Therefore since $b \in R^{+}$ we must have that $L = 0$ .

As an alternative to that, we can find a closed form for $x_{k}$ which allows us to compute the limit directly

Note that in the induction step something interesting occurred, syntactically we could have then replaced $x_{k - 1}$ with $\frac{x_{k - 2}}{1 + b x_{k - 2}}$ then following the same simplification steps we would have obtained $\frac{x_{k - 2}}{1 + 3 b x_{k - 2}}$ , from this we make the conjecture that $x_{j} = \frac{a}{1 + j b a}$ which we will now prove by showing that for any $k \in N_{1}$ , and any $j \in {0, \dots, k}$ we have $x_{k} = \frac{x_{k - j}}{1 + j b x_{k - j}}$

For the base case of $j = 1$ we see that $x_{k} = \frac{x_{k - 1}}{1 + (1) x_{k - 1}}$ , now suppose it holds true for some $j \in {1, \dots, k - 1}$ and we'll show that it holds true for $j + 1$ , $\begin{aligned} x_{k} & = \frac{x_{k - j}}{1 + j b x_{k - j}} \\ = \frac{\frac{x_{k - j - 1}}{1 + b x_{k - j - 1}}}{1 + j b (\frac{x_{k - j - 1}}{1 + b x_{k - j - 1}})} \\ = \frac{\frac{x_{k - j - 1}}{1 + b x_{k - j - 1}}}{1 + (j + 1) \frac{x_{k - j - 1}}{1 + b x_{k - j - 1}}} \\ = \frac{x_{k - j - 1}}{1 + (j + 1) b x_{k - j - 1}} \end{aligned}$

Therefore by finite induction it holds true for all $j \in {1, \dots k}$ specifically for $k = j$ we have $x_{k} = \frac{a}{1 + k b a}$