I think you supposed that $\alpha_n>0$, because if n is even then $f_n(x)$ can't have only one real root. a) If $f_n(x_0)=0$ then $g_n\left(\frac{1}{x_0}\right)=0$, where $g_n(t)=-1+\sum_{k=1}^{n}t^k$. Since $g_n(t)$ is increasing on $\mathbb{R^{+}}$, $g_n(x)$ has exactly one positive root. Hence, $f_n(x)$ has exactly one positive root $\alpha_n$. b) Let $\beta_n=\frac{1}{\alpha_n}>0$. Note that $g_n(\beta_n)=g_{n+1}(\beta_{n+1})=0$. Also, since $\beta_{n+1}>0$, we have $g_n(\beta_{n+1})<g_{n+1}(\beta_{n+1})=g_n(\beta_n)$. Note that $g_n(x)$ is increasing on $\mathbb{R^{+}}$, so $\beta_{n+1}<\beta_n$. From the definition of $\beta_n$ we get that $\alpha_{n}<\alpha_{n+1}$. Therefore, $\{\alpha_n\}$ is a strictly increasing sequence. c) Note that $g_n\left(\frac{1}{2}\right)=-\frac{1}{2^{n}}<g_n(\beta_n)=0$, so $\beta_n>\frac{1}{2}$. From the last we obtain that $\alpha_n<2$. Since $\{\alpha_n\}$ is an increasing sequence, exists $\lim\limits_{n\rightarrow\infty} \alpha_n = A$ and $\lim\limits_{n\rightarrow\infty} \beta_n = B$ ($AB=1$). From $g_n(\beta_n)=0$ we get that $\frac{\beta_n^{n}-1}{\beta_n-1}=\frac{1}{\beta_n}~(*)$. Observe that $\beta_n<\beta_2=\frac{\sqrt{5}-1}{2}<1$, so $\lim\limits_{n\rightarrow\infty}\beta^n_n=0$. When $n$ tends to infinity in $(*)$ we get $B=1-B$, so $B=\frac{1}{2}$. Therefore, $\lim_{n \rightarrow \infty} \alpha_n=2$.

Only one solution? Okay..

$\textbf{Solution (1): }$ A very famous lemma is handy in solving this problem. $\textbf{Lemma:}$ For any polynomial function $f(x)$, if $f(a)\cdot f(b)<0$ for some real numbers $a<b$, then there lies a root of $f(x)=0$ in the interval $[a,b]$. (Note that this works in general whenever $f(x)$ is a continuous function whose domain is a superset of $[a,b]$.) We have $f(1)=1-n < 0$, $f(2)=1\Longrightarrow f(1)\cdot f(2)<0\Longrightarrow$ there is a root $\alpha\in(1,2)$. So, there must always be a positive root of $f_n(x)=0$. So, we only need to prove uniqueness of this root. For the sake of contradiction, let $\alpha\ne\beta$ be positive reals such that $f(\alpha)=f(\beta)=0$. Then, $\alpha^n=\alpha^{n-1}+\cdots+\alpha+1=\frac{\alpha^n-1}{\alpha-1}\Longrightarrow \alpha^{n+1}-2\alpha^n+1=0$. Similarly, $\beta^{n+1}-2\beta^n+1=0$. Subtracting the equations gives $\alpha^{n+1}-\beta^{n+1}=2(\alpha^n-\beta^n)\Longrightarrow\left(\frac{\alpha}{\beta}\right)^{n+1}-2\left(\frac{\alpha}{\beta}\right)^n+1=0$. Let $\frac{\alpha}{\beta}=x$. Since $\alpha\ne\beta$, $x\ne1$, while $x^{n+1}-2x^n+1=0=x^{n+1}-x^n-x^n+1=x^n(x-1)-(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)=(x-1)f_n(x)=0\Longrightarrow f_n(x)=0$. So, if $\alpha$ and $\beta$ are distinct roots, then so is $\frac{\alpha}{\beta},\frac{\alpha}{\beta^2},\cdots$. This gives us infinitely many real roots, but it can have at most $n$ real roots, a contradiction! So, there exists exactly one (unique) positive real root, as required. $\square$ $\textbf{Solution (2): }$ We again use the lemma which was used in proving $(1)$. Note that $f_{n+1}(\alpha_n)={\alpha_n}^{n+1}-{\alpha_n}^n-({\alpha_n}^{n-1}+\cdots+\alpha_n+1)={\alpha_n}^{n+1}-2{\alpha_n}^n={\alpha_n}^n(\alpha_n-2)<0$. So, $f_{n+1}(\alpha_n)f_{n+1}(2)<0$, implying that $\alpha_n<\alpha_{n+1}<2$. Therefore the seqeunce is strictly increasing. $\square$ $\textbf{Solution (3): }$ Define $\gamma_n=\frac{1}{\alpha_n}$. Then, we have $\gamma_n+{\gamma_n}^2+\cdots+{\gamma_n}^n=1$. Moreover, $\frac{1}{2}<\gamma_n<1\Longrightarrow|\gamma_n|<1$ and $(\gamma_n)_n$ is a decreasing. So, $\lim_{n\to\infty}\gamma_n$ exists. We may rewrite our equation as $1={\gamma_n}\cdot\left(\frac{1-{\gamma_n}^n}{1-\gamma_n}\right)$. As $n\to\infty$, ${\gamma_{n}}^n\to0\Longrightarrow 1=\frac{\gamma_n}{1-\gamma_n}\Longrightarrow\gamma_n=\frac{1}{2}\Longrightarrow\alpha_n=2$, as desired. $\square$

Pyramix wrote: $\textbf{Solution (1): }$ A very famous lemma is handy in solving this problem. $\textbf{Lemma:}$ For any polynomial function $f(x)$, if $f(a)\cdot f(b)<0$ for some real numbers $a<b$, then there lies a root of $f(x)=0$ in the interval $[a,b]$. (Note that this works in general whenever $f(x)$ is a continuous function whose domain is a superset of $[a,b]$.) We have $f(1)=1-n < 0$, $f(2)=1\Longrightarrow f(1)\cdot f(2)<0\Longrightarrow$ there is a root $\alpha\in(1,2)$. So, there must always be a positive root of $f_n(x)=0$. So, we only need to prove uniqueness of this root. For the sake of contradiction, let $\alpha\ne\beta$ be positive reals such that $f(\alpha)=f(\beta)=0$. Then, $\alpha^n=\alpha^{n-1}+\cdots+\alpha+1=\frac{\alpha^n-1}{\alpha-1}\Longrightarrow \alpha^{n+1}-2\alpha^n+1=0$. Similarly, $\beta^{n+1}-2\beta^n+1=0$. Subtracting the equations gives $\alpha^{n+1}-\beta^{n+1}=2(\alpha^n-\beta^n)\Longrightarrow\left(\frac{\alpha}{\beta}\right)^{n+1}-2\left(\frac{\alpha}{\beta}\right)^n+1=0$. Let $\frac{\alpha}{\beta}=x$. Since $\alpha\ne\beta$, $x\ne1$, while $x^{n+1}-2x^n+1=0=x^{n+1}-x^n-x^n+1=x^n(x-1)-(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)=(x-1)f_n(x)=0\Longrightarrow f_n(x)=0$. So, if $\alpha$ and $\beta$ are distinct roots, then so is $\frac{\alpha}{\beta},\frac{\alpha}{\beta^2},\cdots$. This gives us infinitely many real roots, but it can have at most $n$ real roots, a contradiction! So, there exists exactly one (unique) positive real root, as required. $\square$ $\textbf{Solution (2): }$ We again use the lemma which was used in proving $(1)$. Note that $f_{n+1}(\alpha_n)={\alpha_n}^{n+1}-{\alpha_n}^n-({\alpha_n}^{n-1}+\cdots+\alpha_n+1)={\alpha_n}^{n+1}-2{\alpha_n}^n={\alpha_n}^n(\alpha_n-2)<0$. So, $f_{n+1}(\alpha_n)f_{n+1}(2)<0$, implying that $\alpha_n<\alpha_{n+1}<2$. Therefore the seqeunce is strictly increasing. $\square$ $\textbf{Solution (3): }$ Define $\gamma_n=\frac{1}{\alpha_n}$. Then, we have $\gamma_n+{\gamma_n}^2+\cdots+{\gamma_n}^n=1$. Moreover, $\frac{1}{2}<\gamma_n<1\Longrightarrow|\gamma_n|<1$ and $(\gamma_n)_n$ is a decreasing. So, $\lim_{n\to\infty}\gamma_n$ exists. We may rewrite our equation as $1={\gamma_n}\cdot\left(\frac{1-{\gamma_n}^n}{1-\gamma_n}\right)$. As $n\to\infty$, ${\gamma_{n}}^n\to0\Longrightarrow 1=\frac{\gamma_n}{1-\gamma_n}\Longrightarrow\gamma_n=\frac{1}{2}\Longrightarrow\alpha_n=2$, as desired. $\square$ Why are $\frac{\alpha}{\beta}$,$\frac{\alpha}{\beta^2}$,... root of polynom when $\alpha$≠$\beta$

I have a different proof for part (a). Since $f_n(1)=1-n<0$ and $f_n(2)=1>0$, it follows by Intermediate Value Theorem that there is a root $\alpha\in(1,2)$. Suppose for contradiction that there are two positive roots, $\alpha$ and $\beta$. WLOG assume that $\alpha<\beta$. We first show that both of them are in $(1,2)$. Note that for $x\in(0,1)$, $0<x^n<x^{n-1}<\cdots<x<1$, so $f_n(x)<0$. Also, $f_n(1)=1-n\ne0$. Moreover, $(x-1)f_n(x)=x^{n+1}-2x^n+1$. So, for $x\in[2,\infty)$, $(x-1)f_n(x)=x^n(x-2)+1\geq1$, so $f_n(x)=0$ is clearly not possible. It follows from above that $1<\alpha<\beta<2$. Note that since $\alpha,\beta\ne1$, it follows $\alpha^{n+1}(2-\alpha)=\beta^{n+1}(2-\beta)=1$. So, $\frac{\alpha^{n+1}}{2-\beta}=\frac{\beta^{n+1}}{2-\alpha}=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}=\alpha^n+\alpha^{n-1}\beta+\cdots+\alpha\beta^{n-1}+\alpha^n$. Using $\alpha<\beta$, we get $n\alpha^n<\frac{\alpha^{n+1}}{2-\beta}=\frac{\beta^{n+1}}{2-\alpha}<n\beta^n\Longrightarrow\frac{\alpha}{2-\beta}>n>\frac{\beta}{2-\alpha}\Longrightarrow(n+1)\beta>\alpha+n\beta>2n>\beta+n\alpha>(n+1)\alpha\Longrightarrow\alpha<\frac{2n}{n+1}<\beta$. We show that there cannot be another root. Suppose there is another root $\gamma$. Consider three cases: $\gamma<\alpha<\beta$, $\alpha<\gamma<\beta$ and $\alpha<\beta<\gamma$. For the first case, we have $\gamma<\frac{2n}{n+1}<\alpha$, but that contradicts with $\frac{2n}{n+1}>\alpha$. Similarly, we get contradictions in the other two cases as well. It follows that: (1) $f_n(x)$ does not change sign in intervals $(0,\alpha)$, $(\alpha,\beta)$, and $(\beta,\infty)$ (and hence $f_n(x)<0$ for $x\in(0,\alpha)$, while $f_n(x)>0$ for $x\in(\beta,\infty)$; (2) $f_n\left(\frac{2n}{n+1}\right)\ne0$. So, the sign of $f_n(x)$ for $x\in(\alpha,\beta)$ is determined by $f_n\left(\frac{2n}{n+1}\right)$. If the sign is positive, then $\beta$ is a double root, if the sign is negative then $\alpha$ is the double root. So in both cases, $f_n(x)=0$ and $f_n'(x)=0$ have a common root, say $r$ (note that $r\in\{\alpha,\beta\}$). Define $g_n(x)=(x-1)f_n(x)=x^{n+1}-2x^n+1$. Then, $g_n'(x)=(n+1)x^n-2nx^{n-1}=f_n'(x)+(x-1)f_n(x)\Longrightarrow g'(r)=f_n'(r)+(r-1)f_n(r)=0=(n+1)r^n-2nr^{n-1}$. Since $r>0$, it follows that $r=\frac{2n}{n+1}$. But this contradicts the fact that $f_n\left(\frac{2n}{n+1}\right)\ne0$. Therefore, the original claim that there are at least two positive roots is wrong. So, there is exactly one (unique) positive root, as required. $\square$ $\text{Remark: }$It can be show that for $n>1$, $f_n\left(\frac{2n}{n+1}\right)<0$.