Problem 6

For every real number

x_1

, define a sequence by

x_{n+1}=x_n\left(x_n+\frac1n\right)

. Prove that there exists exactly one value of

x_1

for which

0<x_n<x_{n+1}<1

for all

n

Step 3 of 5: Choose the common point and verify the bounds

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In plain words

The chosen starting point threads every shrinking gate: after the $n$ th update it is still below $1$ , but already very close to $1$ .

x_1\in\bigcap_{n\ge1}[a_n,b_n]\quad\Longrightarrow\quad 1-\frac1n<x_n<1

Detailed analysis

The nested intervals have a nonempty intersection, so choose $x_1$ in it. It cannot equal any $a_n$ or $b_n$ , so the chosen $x_1$ satisfies $a_n<x_1<b_n$ for every $n$ . Therefore $x_{n+1}=S_n(x_1)$ lies between $1-1/n$ and $1$ . Also $x_n=S_{n-1}(x_1)>S_{n-1}(a_n)=1-1/n$ for $n\ge1$ (with the evident interpretation at $n=1$ ).