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 5 of 5: Prove uniqueness by shrinking the intervals

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In plain words

Any two candidate starts would have to fit inside every interval, but those intervals become narrower than any positive separation; only one start can survive.

b_n-a_n\le\frac{b_n}{n}<\frac1n\longrightarrow0

Detailed analysis

Every valid $x_1$ must satisfy $a_n<x_1<b_n$ for every $n$ , because $x_{n+1}=S_n(x_1)$ lies between $1-1/n$ and $1$ . To show there is at most one such point, note that $S_n$ has nonnegative coefficients and is convex on $[0,b_n]$ . Since $S_n(0)=0$ and $S_n(b_n)=1$ , convexity gives $S_n(x)\le x/b_n$ there. At $x=a_n$ , $1-1/n\le a_n/b_n$ , so $b_n-a_n\le b_n/n<1/n\to0$ . The common intersection therefore has exactly one point.