Our teacher gave us a while ago this problem:
For each positive integer $n$ show that there exists a unique positive real $x_n$ such that: $x_n+\ln(x_n)=n$. Give the first $4$ terms of the expansion of $x_n$.
It is not hard to see that: $x_n=n-\ln(n)+o(\ln(n))$ by some simple bounding arguments, however I couldn’t find the last two. It would be easier if there is some way to systematically find expansion of inverse functions at infinity, and that’s my question, how can we find them?
First of all, we will show that for each positive integer $\,n\,$ there exists a unique positive real $\,x_n\,$ such that $\;x_n+\ln x_n=n\;.$
Let $\;f:(0,+\infty)\to\Bbb R\;$ be the function defined as follows
$f(x)=x+\ln x-n\;\;\;$ for any $\;x\in(0,+\infty)\,.$
Since the function $\,f\,$ is continuous on $\,(0,+\infty)\;,$ $\,f(1)=1-n\leqslant0\;$ and $\;f(n)=\ln n\geqslant0\;,\;$ there exists $\,x_n\in[1,n]\,$ such that $\,f(x_n)=0\,.$
Moreover $\,x_n\,$ is the unique solution of the equation $\,f(x)=0\,$ because the function $\,f\,$ is monotonically increasing on $\,(0,+\infty)\,,\,$ indeed $\,f\,$ is differentiable on $\,(0,+\infty)\,$ and $\,f’(x)=1+\frac1x>0\,,\,$ for any $\,x>0\,.$
Consequently, for each positive integer $\,n\,$ there exists a unique positive real $\,x_n\in[1,n]\,$ such that $\;x_n+\ln x_n=n\;.$
Since $\,x_n\in[1,n]\;$ and $\;x_n=n-\ln x_n\,,\,$ for any $\,n\in\Bbb N\,,\,$ it follows that $\;1-\dfrac{\ln n}n\leqslant\dfrac{x_n}n=1-\dfrac{\ln x_n}n\leqslant1\;$ for any $\,n\in\Bbb N$
and, by applying the Squeeze Theorem, we get that $\;\lim\limits_{n\to\infty}\dfrac{x_n}n=1\;,\;$ hence, $\;x_n\sim n\;$ as $\;n\to\infty\;,$
or equivalently, $\;\,x_n=n+o(n)\,.$
Now we will get the first $\,4\,$ terms of the expansion of $\,x_n\,.$
$\begin{align}x_n&=n\!-\!\ln x_n=n\!-\!\ln\left(n\!-\!\ln x_n\right)=n\!-\!\ln\!\big[n\!-\!\ln\left(n\!-\!\ln x_n\right)\big]\!=\\[3pt]&=n\!-\!\ln\!\big\{n\!-\!\ln\!\big[n\!-\!\ln\!\big(n\!+\!o(n)\big)\big]\big\}\!=\\[3pt]&=n\!-\!\ln\!\big\{n\!-\!\ln\!\big[n\!-\!\ln n\!-\!\ln\!\big(1\!+\!o(1)\big)\big]\big\}\!=\\[3pt]&=n\!-\!\ln\!\big\{n\!-\!\ln\!\big[n\!-\!\ln n\!-\!o(1)\big]\big\}\!=\\[3pt]&=n\!-\!\ln\!\left\{n\!-\!\ln n\!-\!\ln\!\left[1\!-\!\dfrac{\ln n}n\!-\!o\!\left(\!\dfrac1n\!\right)\right]\right\}\!=\\[3pt]&=n\!-\!\ln\!\left\{n\!-\!\ln n\!+\!\dfrac{\ln n}n\!+\!o\!\left(\!\dfrac1n\!\right)\!\right\}\!=\\[3pt]&=n\!-\!\ln n\!-\!\ln\!\left\{1\!-\!\dfrac{\ln n}n\!+\!\dfrac{\ln n}{n^2}\!+\!o\!\left(\!\dfrac1{n^2}\!\right)\!\right\}\!=\\[3pt]&=n\!-\!\ln n\!+\!\dfrac{\ln n}n\!-\!\dfrac{\ln n}{n^2}\!-\!o\!\left(\!\dfrac1{n^2}\!\right)\!+\!\dfrac12\left(\!\dfrac{\ln n}n\!\right)^{\!\!2}\!\!+\!O\!\left(\!\dfrac{\ln^3\!n}{n^3}\!\right)\!=\\[3pt]&=n\!-\!\ln n\!+\!\dfrac{\ln n}n\!-\!\dfrac{\ln n}{n^2}\!+\!\dfrac{\ln^2\!n}{2n^2}\!+\!o\!\left(\!\dfrac1{n^2}\!\right).\end{align}$
So the first $\,4\,$ terms of the expansion of $\,x_n\,$ are :
$x_n=n-\ln n+\dfrac{\ln n}n+\dfrac{\ln^2\!n-2\ln n}{2n^2}+o\!\left(\!\dfrac1{n^2}\!\right).$