MathLabs

Problem 6

For every real number x1x_1, define a sequence by xn+1=xn(xn+1n)x_{n+1}=x_n\left(x_n+\frac1n\right). Prove that there exists exactly one value of x1x_1 for which 0<xn<xn+1<10<x_n<x_{n+1}<1 for all nn.
Step 4 of 5: Obtain strict increase
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In plain words

Being above the fixed threshold makes the next multiplier exceed 11, so each update grows the current positive value; the upper gate keeps it from crossing 11.

xn+1−xn=xn(xn+1n−1)>0x_{n+1}-x_n=x_n\left(x_n+\frac1n-1\right)>0
Detailed analysis

From Step 3, xn>1−1/nx_n>1-1/n and xn>0x_n>0. Hence xn+1/n>1x_n+1/n>1, and the recurrence gives xn+1=xn(xn+1/n)>xnx_{n+1}=x_n(x_n+1/n)>x_n. Step 3 also gives xn+1<1x_{n+1}<1, so 0<xn<xn+1<10<x_n<x_{n+1}<1 for every nn.