$f : \mathbb R \to \mathbb R, f(x) = x – 1 + e^{-x}$. Let sequence $(x_n)_{n \in \mathbb N} , x_0 \ge 0, x_n = f(x_{n-1})$. Prove $x_n$ is convergent.

Question

$f : \mathbb R \to \mathbb R, f(x) = x – 1 + e^{-x}$.
Let sequence $(x_n)_{n \in \mathbb N} , x_0 \ge 0, x_n = f(x_{n-1})$. Prove $x_n$ is convergent and calculate $\lim_{n \to \infty}x_n$.

I am really disheartened, I just realized that I learned nothing in the past year that I tried to better myself at math, I don't know what to do, I keep asking questions here, get complicated answers, and the next time I come up across the same exercise I find I know absolutely nothing about it.

I understand what the exercise asks me to prove but I had no clue why it was true until I plotted the graph on desmos. In my mind, I thought this sentence was impossible, but only because I suck at math apparently. I realized, seeing the graph, that for any input $x \in (0, \infty)$ the output will be less than the input. So $f(x) < x$ for any $ x \ge 0$.

I would have never realized that by myself, and even now that I know it, I have no clue how to prove $x_n$ is convergent. Help me, please, I am losing my mind.

Answer 1

Simpler way

Convergence can be obtained via the monotone convergence theorem by noting that the sequence $(x_n)$ is decreasing and bounded below by $0$ for any $x_0\geq 0$.

To prove that for any choice of $x_0$ the corresponding sequence $(x_n)$ converges to zero, one may argue as follows. We have already that $(x_n)$ converges, hence $(x_n)$ is Cauchy. It follows that $x_{n+1} - x_n \overset{n}{\longrightarrow} 0$. But $\displaystyle{x_{n+1}-x_n = \dfrac{1}{\mathrm{e}^{x_n}} - 1}$, completing the proof.

Edit (answer to comment). To show that $(x_n)$ is decreasing and bounded below by $0$, it suffices to prove that for any positive $x$ the inequalities $0 < f(x) < x$ hold. Can you see why it suffices to prove this? (Hint: think of $x$ as the initial $x_0$). Try to also prove the inequalities.

Harder way (more generally applicable)

One may apply the unique fixed point theorem for the ordinary numerical iteration method, that is $x_n =f(x_n), n\in \mathbb{N}_+$. (Banach's fixed point theorem works just as well). Let $x_0>0$ be given. In order to apply the theorems in this case, one should first show that our $f$ is a contraction map on $[0, x_0]$ (continuity of $f$ is clear, and $[0, x_0]$ is complete). For this, it suffices to show that there exists a $c\in[0, 1)$ such that $|f'| < c$ in $[0, x_0]$.

But we have, for any positive real $x$, that $f'(x) = 1 - 1/\mathrm{e}^{x}$. This is strictly increasing for positive $x$, therefore $0\leq f'(x) < 1 - 1/\mathrm{e}^{x_0} =: c < 1$ in $[0, x_0]$. Note also that $f(0) = 0$ (in other words, $0$ is a fixed point of $f$ on $[0, x_0]$).

All that remains to show is $f([0, x_0]) \subseteq [0, x_0]$. Can you do this, completing the proof?