Let $P_n(x)=x^n-nx+1$ be a sequence of polynomials, where $P_n\colon[1, +\infty) \to \mathbb{R}$ and $n \ge 2$
a) Show that for each $n$, $P_n(x)=0$ has exactly one solution, and for each $n$ let $x_n$ be that unique number such that $P_n(x_n)=0$.
b)Show that $\lim_{n \to \infty}x_n=1$ and study the convergence of the series $\sum_{n \ge 2} (x_n-1)^{\alpha}$, where $\alpha \in \mathbb{R}$
My approach:
For once $\frac{dP_n(x)}{dx}=n(x^{n-1}-1) \ge 0, \forall x \ge 1$, so $P_n$ is increasing. $P_n(1)=2-n \le 0$ and $\lim_{x \to +\infty}P_n(x)=\infty$, so by the intermediate value property there is at least one point $x_n$, where $P_n(x_n)=0$ and it is unique by monotonicity.
Now for the limit.
Take $b \in (0, 1]$. Calculate $$P_n\left(1+\frac{1}{n^b}\right)=\left(1+\frac{1}{n^b}\right)^n-n\left(1+\frac{1}{n^b}\right)+1=2-n+\sum_{k=2}^n {n\choose k}\frac{1}{n^{kb}} \ge 2-n + \sum_{k=2}^m{n\choose k}\frac{1}{n^{kb}}$$
where $m \le n$. The left side is a polynomial in $n$. The highest power of $n$ with a positive coeficient is $n^{m-mb}$ and the one with a negative coefficient is $n$. If we want the limit to be positive, we want $m-mb\ge 1 \iff b \le 1-\frac{1}{m}$ which is true for sufficiently large $m$ so for sufficiently large $n$. Therefore $\forall b \in (0, 1)\colon P_n(1+\frac{1}{n^b}) \ge 0 \iff x_n < 1 + \frac{1}{n^b}$.
Now calculate $$P_n\left(1+\frac{1}{n}\right)=\left(1+\frac{1}{n}\right)^n-n\left(\frac{n+1}{n}\right)+1=\left(1+\frac{1}{n}\right)^n-n$$. So the right hand goes to $-\infty$ so for a sufficiently large $n,P_n(1+\frac{1}{n}) \leq 0 \iff 1+\frac{1}{n} \leq x_n$.
Now we have the double inequality $\forall b \in (0, 1)\colon \frac{1}{n} \leq x_n - 1 \leq \frac{1}{n^b}$ so by the Squeeze theorem $\lim_{n \to \infty}x_n=1$.
Now finding whether or not the series converges. From the above inequality $\frac{1}{n^\alpha} \leq (x_n-1)^\alpha$ so for $\alpha \leq 1$ the series diverges.
To take care of the case of $\alpha \geq 1$, I will use the following lemma.
Let $L^*=\limsup_{n \to \infty}\frac{u_n}{v_n} \in [0, +\infty)$, where $u_n, v_n \geq 0$. Then $\sum v_n \text{converges} \implies \sum u_n \text{converges}$.(There is a similar theorem for $\liminf$ but it is not needed here
Proof:
For some fixed $\epsilon$ we have that $\frac{u_n}{v_n} \leq \sup_{m \geq n}\frac{u_m}{v_m} \leq L^*+\epsilon \implies u_n \leq (L^*+\epsilon)v_n$ and the result follows by direct comparison.
Back to our problem, we have that $$\forall b \in (0, 1)\colon n^{b-1} \leq n^b(x_n-1) \leq 1 \implies \limsup_{n \to \infty}n^b(x_n-1) \in [0, 1] \implies \limsup_{n \to \infty} \frac{x_n-1}{\frac{1}{n^b}} = \limsup\frac{(x_n-1)^\alpha}{n^{b\alpha}} \in [0, +\infty)$$ So $\sum_{k=2}^n(x_n-1)^\alpha \text{converges if }\sum_{k=2}^n\frac{1}{n^{b\alpha}}$, the latter converges if $b\alpha > 1 \iff b > \frac{1}{\alpha}$. So for a given(fixed) $\alpha > 1$, we can always choose $b > \frac{1}{\alpha}$. So the series of $(x_n-1)^\alpha$ converges if $\alpha>1$ and diverges otherwise.
This problem was from some sort of contest and I wonder how would it be appropriate to approach a problem like this. The inequalities for $x_n$ make the problem seem easier, but it took me 2-3hours to find them, which is not feasible to do under the time of the contest, keeping in mind you have 4 problems of the same-ish difficulty. While the limit of $x_n$ could be calculated more easily(firstly $x_n$ is obviously decreasing and is bounded bellow by $1$ so it is convergent by Weierstrass and from the expression $x_n=(nx_n-1)^\frac{1}{n}$ and now we apply Cauchy D'Alambert, the root criterion. So $\lim\frac{(n+1)x_{n+1}-1}{nx_n-1}=1=\lim x_n$ ), the series is atrocious to calculate without first finding the appropriate inequalities. How would one approach problems like this?
P.S.: Let me know if my proof is correct.
