Root in $(1,2]$ of Equation $x^n-x-n=0$ Consider the equation $x^n-x-n=0$ with $n\in\mathbb{N},n\geq2.$
$a)$ Show that this equation has exactly one solution $u_n\in(1,2].$
$b)$ Show that the sequence $\left\{u_n\right\}$ is decreasing.
$c)$ Determine $L=\lim_{n\rightarrow\infty}u_n.$
Here is what I've done so far on this problem.

$a)$
Let $f_n(x)=x^n-x-n,$ then clearly $f_n$ is continuous and $f_n(1)=-n<0.$
Let $g(n)=f_n(2)=2^n-2-n,$ then $g'(n)=2^n\cdot\ln2-1>0$ as $n\geq2,$ hence
$$f_n(2)=g(n)\geq g(2)=2^2-2-2=0.$$
Furthermore, $\frac{d}{dx}f_n(x)=nx^{n-1}-1\geq1-1=0$ as $n\geq2$ and $x\geq1.$
Therefore, $f_n$ is non$-$decreasing on $[1,2]$ and $f_n(1)<0\leq f_n(2),$ and the Intermediate Value Theorem implies the unique root $u_n$ as desired.


$c)$ Note that $u_n^n-u_n-n=0\Leftrightarrow u_n=\sqrt[n]{u_n+n}$ and for $a>0$ fixed,
$$\lim_{n\rightarrow\infty}\ln(\sqrt[n]{a+n})=\lim_{n\rightarrow\infty}\frac{\ln(a+n)}{n}=\lim_{n\rightarrow\infty}\frac{1}{a+n}=0$$
hence
$$1=\lim_{n\rightarrow\infty}\sqrt[n]{1+n}\leq\lim_{n\rightarrow\infty}\sqrt[n]{u_n+n}\leq\lim_{n\rightarrow\infty}\sqrt[n]{2+n}=1$$
so $L=1.$

I'm currently stuck with part $b),$ so any hints/ideas/comments are appericated. Thank you!
 A: An alternative approach: let $p_n(x)=x^n-x-n$. We may notice that
$$ p_n\left(1+\frac{\log(n+1)}{n}\right)=\left(1+\frac{\log(n+1)}{n}\right)^n-(n+1)-\frac{\log(n+1)}{n} < -\frac{\log(n+1)}{n} $$
so $u_n > 1+\frac{\log(n+1)}{n}$. Since $p_n(x)$ is convex, an upper bound for $u_n$ can be obtained by applying one step of Newton's method with starting point $1+\frac{\log(n+1)}{n}$, leading to an alternative proof of point (c).
$$ p_{n+1}(u_n) = u_n\cdot u_n^n - u_n - (n+1) = u_n\cdot(u_n+n)-u_n-(n+1) = u_n^2+(n-1)u_n-(n+1)$$
and if we prove $p_{n+1}(u_n)>0$ point (b) is easily done by induction. $p_{n+1}(u_n)>0$ is equivalent to
$$ u_n > \frac{1-n+\sqrt{n^2+2n+5}}{2} = 1+\frac{2}{n+1+\sqrt{n^2+2n+5}}$$
and
$$ \frac{\log(n+1)}{n} > \frac{1}{n+1}>\frac{2}{n+1+\sqrt{n^2+2n+5}} $$
holds for any $n\geq 2$.
A: First, I shall prove a stronger claim.

Claim: $u_n \geq 1 + \frac{2}{n}$ for $n \geq 2$.
Proof: Applying IVT on $\left(1 + \frac{2}{n}, 2\right)$, we simply have to evaluate $f_n\left(1 + \frac{2}{n}\right)$. Now,
$$
f_n\left(1 + \frac{2}{n}\right) = \left(1 + \frac{2}{n}\right)^n - \left(1 + \frac{2}{n}\right) - n
$$
For $n \geq 7$, notice that $f_n\left(1 + \frac{2}{n}\right) \leq e^2 - (n + 1)$, so the inequality holds. For $2\leq n\leq 6$, simply substitute the values in.

Claim: $\frac{\sqrt{n^2 + 4} - (n - 2)}{2} \leq 1 + \frac{2}{n}$.
Proof: The inequality is equivalent to
$$\begin{align*}
n\sqrt{n^2 + 4} - n(n - 2) &\leq 2n + 4 \\
n\sqrt{n^2 + 4} &\leq n^2 + 4 \\
n^2(n^2 + 4) &\leq (n^2 + 4)^2 \\
4(n^2 + 4) \geq 0
\end{align*}$$
which holds for all $n > 0$.

Now, Apply IVT on $(1, u_{n - 1})$. In particular, let $u = u_{n - 1}$. Then,
$$\begin{align*}
u^{n - 1} - u - (n - 1) = 0 &\implies u^{n - 1} = u + (n - 1) \\\\
u^n - u - n &= u(u + (n - 1)) - u - n \\
&= u^2 + (n - 2)u - n
\end{align*}$$
Now, note that this quadratic has positive root
$$
u = \frac{-(n - 2) + \sqrt{(n - 2)^2 + 4n}}{2} = \frac{-(n - 2) + \sqrt{n^2 + 4}}{2} \leq 1 + \frac{2}{n}
$$
Since $u \geq 1 + \frac{2}{n}$, we have that $f(u) = u^n - u - n > 0$. Clearly, $f(1) = -n < 0$, so applying IVT gives part (b).
