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 1 of 5: Encode the recurrence as iterated maps

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In plain words

Instead of following one starting value through many steps, package the first $n$ updates into one increasing curve $S_n$ .

S_0(t)=t,\qquad S_n(t)=S_{n-1}(t)\left(S_{n-1}(t)+\frac1n\right)

Detailed analysis

Define $S_0(t)=t$ and recursively $S_n(t)=S_{n-1}(t)(S_{n-1}(t)+1/n)$ . Then $x_{n+1}=S_n(x_1)$ . Each $S_n$ is a polynomial with nonnegative coefficients, hence is continuous and strictly increasing on $[0,1]$ ; also $S_n(0)=0$ and $S_n(1)>1$ .