Let \(0<x_1<1\) and define \[ x_{n+1}=x_n-x_n^{n+1}. \] Show that \[ \liminf_{n\to\infty}x_n>0. \]
Solution
Set \[ q\coloneqq x_1, \] so \[ 0<q<1. \]
1. For every \(n\geq1\),\[ 0<x_{n+1}<x_n\leq q<1. \]
Suppose \(0<x_n<1\). Then \[ 0<x_n^n<1, \] so \[ 0<1-x_n^n<1. \] The recurrence can be written \[ x_{n+1} = x_n(1-x_n^n), \] which therefore satisfies \[ 0<x_{n+1}<x_n. \] Starting from \(0<x_1=q<1\), induction proves the assertion.
2. For every \(n\geq2\),\[ x_n = q\prod_{k=1}^{n-1}(1-x_k^k). \]
The recurrence gives \[ \frac{x_{k+1}}{x_k} = 1-x_k^k \] because step 1 ensures \(x_k\neq0\). Multiplying these identities for \(k=1,\ldots,n-1\) telescopes to the displayed product.
3. For every \(n\geq2\),\[ x_n \geq q\prod_{k=1}^{n-1}(1-q^k). \]
By step 1, \[ 0<x_k\leq q<1. \] Hence \[ x_k^k\leq q^k, \] so \[ 1-x_k^k \geq 1-q^k. \] Apply this factor by factor in the product of step 2.
4. The infinite product\[ P\coloneqq \prod_{k=1}^{\infty}(1-q^k) \]converges to a strictly positive number.
Choose \(K\) so large that \[ q^k\leq\frac12 \] for every \(k\geq K\). For \(0\leq t\leq1/2\), \[ \log(1-t)\geq-2t. \] Therefore \[ \sum_{k=K}^{\infty}\log(1-q^k) \geq -2\sum_{k=K}^{\infty}q^k > -\infty. \] The logarithmic series also consists of nonpositive terms, so its partial sums decrease to a finite real limit. Exponentiating shows that \[ \prod_{k=K}^{\infty}(1-q^k)>0. \] The finitely many factors with \(k<K\) are all positive, hence the complete product \(P\) is positive.
5. For every \(n\),\[ x_n\geq qP>0. \]
Each factor \(1-q^k\) lies in \((0,1)\). Thus the finite products \[ \prod_{k=1}^{n-1}(1-q^k) \] decrease to the infinite product \(P\) and are therefore at least \(P\). Step 3 gives \[ x_n \geq q\prod_{k=1}^{n-1}(1-q^k) \geq qP. \] The case \(n=1\) also satisfies \(x_1=q\geq qP\) because \(P\leq1\).
6. One has\[ \boxed{ \liminf_{n\to\infty}x_n \geq qP >0 }. \]
Step 5 gives the same positive lower bound for every term of the sequence. Taking the lower limit preserves that inequality.
7. Q.E.D.
Step 6 is the required conclusion.