Use the Monotone Convergence Theorem to show the sequence $(s_n)_{n\ge0}$ converges to a limit

Question

Let $(s_n)_{n\ge0}$ be defined by $s_0=1$, $s_{n+1}=f(s_n)$ where $f(x)=x-\frac{x^9}{9}$.

I started by trying to show the sequence was monotonic decreasing using induction so I showed the base case: $s_0=1$, $s_1=1-\frac{1^9}{9}=\frac{8}{9}$ $\rightarrow$$s_0>s_1$. I then assumed $s_n<s_{n-1}$ for all $n<k$ and attempted to show that $s_{k+1}<s_k$.

From our inductive step, we have that $$ s_k<s_{k-1}<1 $$ $$ [s_k-\frac{(s_k)^9}{9}]<1-\frac{(s_k)^9}{9} $$ $$ s_{k+1}<1-\frac{(s_k)^9}{9} $$ Now since $s_k<1$ can I sub that into this somehow. I am confused on how to finish this half of the proof, however, I see how from this we can easily show that the function must then also be bounded by $0$. Any help would be appreciated!

Also, would it have been easier to first show that the sequence is bounded below by 0. Because if we knew that, then we could also just as easily show by induction that the sequence would be monotonic decreasing. Is there a general rule as to which one is easier to show first?

Answer 1

First, show using induction that sequence is contained in $[0,1]$. Base case is trivial and so is inductive step. Now you know your sequence is bounded on both sides and you can show using induction, the way you wanted, that sequence is decreasing. We have $s_{n+1}-s_n=s_n-\frac{s_n^9}{9}-s_n=-\frac{s_n^9}{9}\leq 0$

Answer 2

Just change the claim to $0 < s_{k+1} < s_k < 1$, for all $k \geq 1$. Doing the inductive step: $$ s_{k+1} = f(s_k) = s_k - \frac{(s_k)^9}{9} < s_k + 0 = s_k,\ \text{since $0 < s_k$}. $$ Also, notice that $$ s_{k+1} = s_k \bigg(1 - \frac{s_k^8}{9}\bigg) > 0, $$ since it is the product of two positive numbers.

Answer 3

Show $f$ is monotonically increasing for $0\leq x\leq 1$ and note that $$s_{n} \leq s_{n-1} \Rightarrow s_{n+1} = f(s_n) \leq f(s_{n-1}) = s_n. $$ Since the sequence is bounded from below, $s_n \xrightarrow[n\to\infty]{} \inf \{s_n \mid n\in\mathbb N\}$.