Find the equivalent infinitesimal of $1-x_n$ Suppose $\{x_n\}$ is a sequence satisfying $x_n^n +x_n=1$ . Prove that
$$1-x_n \sim \frac{\ln n}{n}$$
I've shown that $x_n>1-\frac{\ln n}{n}$ by noticing
$$\left(1-\frac{\ln n}{n}\right)^n + 1-\frac{\ln n}{n} = e^{n\ln\left(1-\frac{\ln n}{n}\right)} + 1-\frac{\ln n}{n}<1+\frac{1-\ln n}{n}<1$$
I wonder if there is a $k_n\to1$, such that $x_n<1-\frac{k_n\ln n}{n}$ , but I can't figure out how to construct that.
Thanks so much.
 A: We can show $1- x_n > \frac{\ln n}{2n} \iff x_n < 1-\frac{\ln n}{2n}$. And, this will prove $1-x_n \backsim \frac{\ln n}{n}$ combined with what you've already shown.
Let $f(x) = x^n + x -1$. Then, we want to show $f(1-\frac{\ln n}{2n})>0$. To do so, note that, for any $C>1$, $\ln (1-x) > -Cx$ holds for sufficiently small $x>0$. So, by taking $n$ sufficiently large, we have $\ln (1-\frac{\ln n}{2n})>-C\frac{\ln n}{2n}$.
From this we can obtain a lower bound of $f(1-\frac{\ln n}{2n})$ as follows:
\begin{array}
\, f(1-\frac{\ln n}{2n}) & = \exp \{n\ln (1-\frac{\ln n}{2n}) \} - \frac{\ln n}{2n} \\
& > \exp (-nC\frac{\ln n}{2n}) - \frac{\ln n}{2n} \\
& = \frac{1}{n^{C/2}} - \frac{\ln n}{2n}
\end{array}
If we choose $1<C<2$ then the last amount is positive for sufficiently large $n$.
Note also that the number '$2$' is nothing special. For any $D>1$ choose $C$ so that $1<C<D$. Then, you can show $f(1-\frac{\ln n}{Dn}) > \frac{1}{n^{C/D}} - \frac{\ln n}{Dn} > 0$ for sufficiently large $n$. This implies there is a $k_n \to 1$ as you desire.
