Question 26
S4 (NZL). Suppose that $x_1,x_2,x_3,\ldots$ are positive real numbers for which
$$x_n^n=\sum_{j=0}^{n-1}x_n^j$$
for $n=1,2,3,\ldots$. Prove that for all $n$,
$$2-\frac1{2^{n-1}}\le x_n\lt 2-\frac1{2^n}.$$
$$x_n^n=\sum_{j=0}^{n-1}x_n^j$$
for $n=1,2,3,\ldots$. Prove that for all $n$,
$$2-\frac1{2^{n-1}}\le x_n\lt 2-\frac1{2^n}.$$