Asymptotic expansion of $u_n$ such that $u_n^n-u_n-n=0$

Question

One can show that that the function $f_n : x \mapsto x^n-x-n$ admits a unique root (if $x\ge 1$). Then, we define $u_n\in [1,+\infty)$ such that $u_n^n-u_n-n=0$. I can prove that $\displaystyle \lim_{n\to +\infty} u_n =1$. Then I set $u_n=1+\epsilon_n$ (with $\displaystyle \lim_{n\to +\infty} \epsilon_n =0$) and I've tried to find an equivalent of $\epsilon_n$ (when $n \to +\infty$) as follows $$ (1+\epsilon_n)^n = 1+\epsilon_n+n \qquad \Longrightarrow \qquad n \ln(1+\epsilon_n)=\ln(n+1+\epsilon_n)$$ Now since $\displaystyle \lim_{n\to +\infty} \epsilon_n =0$, I know that $n \ln(1+\epsilon_n) \underset{n\to +\infty}{\sim} n\epsilon_n$ and $\ln(n+1+\epsilon_n)\underset{n\to +\infty}{\sim} \ln(n)$ which gives me $$ \epsilon_n \underset{n\to +\infty}{\sim} \frac{\ln(n)}{n}$$ But "numerically", it seems wrong. Where is my mistake?

Answer 1

Looks good to me! I find the same result, and numerical agreement. Here's a plot of the ratio of the solution to your approximation, $\epsilon_n / (\ln n/n)$:

This is indeed tending fairly rapidly to $1$.

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The next corrections to the equation are $-n \epsilon^2/2 \sim (\ln n)^2/n$ on the LHS and $1/n$ on the RHS. Thus the LHS correction is larger. So let's solve $n \epsilon - n \epsilon^2/2 \sim \ln n$. This gives the next correction: $$\epsilon_n \sim \frac{\ln n}{n} + \frac{(\ln n)^2}{2n^2} + \cdots$$ This indeed matches with the error too, if we plot $(\epsilon_n - \ln n/n)/((\ln n)^2/2n^2)$: