$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

For the induction step we'll notice before anything else that in general for any $n\in {\mathbb{N}}_1$ we have that $$\begin{array}{cccc}& x_{n+1}& ={(x_n^{-1}+b)}^{-1}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{1}{\frac{1}{x_n}+b}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{1}{\frac{1+b{x}_n}{x_n}}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{x_n}{1+b{x}_n}& \text{<span class="tml-eqn"></span>}\end{array}$$ moving on, let $k\in {\mathbb{N}}_1$ we'll prove that $x_k\ge x_{k+1}$, $$\begin{array}{cccc}& x_{k+1}& =\frac{x_k}{1+b{x}_k}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{\frac{x_{k-1}}{1+b{x}_{k-1}}}{1+b\left(\frac{x_{k-1}}{1+b{x}_{k-1}}\right)}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{\frac{x_{k-1}}{1+b{x}_{k-1}}}{\left(\frac{1+b{x}_{k-1}+b{x}_{k-1}}{1+b{x}_{k-1}}\right)}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{x_{k-1}}{1+2b{x}_{k-1}}& \text{<span class="tml-eqn"></span>}\end{array}$$ But then because of the fact that $1+2b{x}_k>1+b{x}_k$ then we know that $$x_{k+1}=\frac{x_{k-1}}{1+2b{x}_{k-1}}<\frac{x_{k-1}}{1+b{x}_{k-1}}=x_k$$

Finally as $x_0=a>\frac{a}{1+ab}=x_1$ as $ab\in {ℝ}^{+}$, then we know that $x_k>x_{k+1}$ for all $k\in {\mathbb{N}}_0$

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 $$\begin{array}{cccc}& \underset{n\to \infty }{\lim }{x}_n& =L& \text{<span class="tml-eqn"></span>}\\ & & =\underset{n\to \infty }{\lim }{x}_{n+1}& \text{<span class="tml-eqn"></span>}\\ & & =\underset{n\to \infty }{\lim }\frac{1}{\frac{1}{x_n}+b}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{1}{\frac{1}{L}+b}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{L}{1+Lb}& \text{<span class="tml-eqn"></span>}\end{array}$$ We see that this implies $1+Lb=1$ Therefore since $b\in {ℝ}^{+}$ we must have that $L=0$.

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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 $\frac{x_{k-2}}{1+3b{x}_{k-2}}$, from this we make the conjecture that $x_j=\frac{a}{1+jba}$ which we will now prove by showing that for any $k\in {\mathbb{N}}_1$, and any $j\in \{0,\dots ,k\}$ we have $$x_k=\frac{x_{k-j}}{1+jb{x}_{k-j}}$$

For the base case of $j=1$ we see that $x_k=\frac{x_{k-1}}{1+\left(1\right)x_{k-1}}$, now suppose it holds true for some $j\in \{1,\dots ,k-1\}$ and we'll show that it holds true for $j+1$, $$\begin{array}{cccc}& x_k& =\frac{x_{k-j}}{1+jb{x}_{k-j}}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{\frac{x_{k-j-1}}{1+b{x}_{k-j-1}}}{1+jb\left(\frac{x_{k-j-1}}{1+b{x}_{k-j-1}}\right)}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{\frac{x_{k-j-1}}{1+b{x}_{k-j-1}}}{1+(j+1)\frac{x_{k-j-1}}{1+b{x}_{k-j-1}}}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{x_{k-j-1}}{1+(j+1)b{x}_{k-j-1}}& \text{<span class="tml-eqn"></span>}\end{array}$$

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