India_2017

(a) Since $f_n(x)$ has one change of sign, it follows that $f_n(x)=0$ has at most one positive real root by Descartes' Rule. As $f_n(0)=-1<0$ and $f_n(2)=2^n-2^{n-1}-2^{n-2}-\cdots -2-1=1>0$, we see that $f_n(x)=0$ has at least one root between $0$ and $2$. Thus $f_n(x)=0$ has exactly one root ${\alpha }_n$ between $0$ and $2$, $n\ge 2$.

(b) We show that ${\alpha }_n<{\alpha }_{n+1}$, for $n\ge 2$. Because $f_n({\alpha }_n)=0$, we have ${\alpha }_n^n-{\alpha }_n^{n-1}-{\alpha }_n^{n-2}-\cdots -{\alpha }_n-1=0$. Now $f_{n+1}({\alpha }_n)={\alpha }_n({\alpha }_n^n-{\alpha }_n^{n-1}-{\alpha }_n^{n-2}-\cdots -{\alpha }_n-1)-1={\alpha }_n\cdot 0-1=-1<0$. As $f_{n+1}(2)=1>0$, as above, we infer that there is a root of $f_{n+1}(x)=0$ between ${\alpha }_n$ and $2$. Since ${\alpha }_n>0$, we conclude that ${\alpha }_n<{\alpha }_{n+1}<2$, as desired.

(c) Again $$(x-1)f_n(x)=(x-1)(x^n-x^{n-1}-x^{n-2}-\cdots -x-1)=x^{n+1}-2x^n+1.$$ So $${\alpha }_n^{n+1}-2{\alpha }_n^n+1=0,$$ giving ${\alpha }_n=2-\frac{1}{{\alpha }_n^n}\ge 2-\frac{1}{{\alpha }_2^n}=\frac{1+\sqrt{5}}{2}$, where ${\alpha }_2=\frac{1+\sqrt{5}}{2}$, being the positive root of $x^2-x-1=0$. But ${\alpha }_2>1$. So $\frac{1}{{\alpha }_2^n}\to 0$, as $n\to \infty$. Thus, in $2-\frac{1}{{\alpha }_2^n}\le {\alpha }_n<2$, we let $n\to \infty$, to get $$\underset{n\to \infty }{\lim }{\alpha }_n=2.$$