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 2 of 5: Trap the initial value in intervals

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In plain words

Each interval consists of starting values that land between a moving lower target and the ceiling $1$ after $n$ updates; the intervals tighten from both sides.

S_n(a_n)=1-\frac1n,\qquad S_n(b_n)=1,\qquad a_n<b_n

The map

u\mapsto u(u+1/2)

used at the first nontrivial fixed-point step.

Detailed analysis

By continuity and strict increase, there are unique $a_n,b_n\in(0,1)$ with $S_n(a_n)=1-1/n$ and $S_n(b_n)=1$ . The map $u\mapsto u(u+1/n)$ fixes $u=1-1/n$ , so from the first equation we also have $S_{n-1}(a_n)=1-1/n$ . Comparing the recursion at these boundary points gives $a_n<a_{n+1}$ and $b_n>b_{n+1}$ . Thus the closed intervals $[a_n,b_n]$ are nested.