For $n \geq 2$, show that $\exists$ a number $\theta_n, \theta_n > 1$ such that $-\log(1-\frac{1}{n}) = \frac{1}{n} + \frac{\theta_n}{2n^2}$
$\lim_{n\to \infty} \theta_n$
My attempt: I am not exactly sure what to do for # 1. I am stuck after doing this... ${\theta_n}=2n^2(-\log(1-\frac{1}{n}) - \frac{1}{n})$
For # 2, I made this real messy by taking derivative and applying L'hopital's rule and such. So, basically this is what I did:
${\theta_n}=2n^2(-\log(1-\frac{1}{n}) - \frac{1}{n}) = -2n^2 \log(1-\frac{1}{n}) - 2n$
If we take limit, then
$\lim_{n\to \infty} \theta_n = \infty$. So, we take derivative to apply L'hopital's:
$\lim_{n\to \infty} \theta_n = - 4n\cdot\log(1-\frac{1}{n}) -\frac{2n}{n-1} - 2 = \infty$.
$\lim_{n\to\infty} \theta_n = -4\log(1-\frac{1}{n}) - \frac{4n}{(n-1)^3}$
I am lost here as this is getting more and more complicated...Any help? Thanks.
For 1), using Taylor's theorem with Lagrange's remainder for $\log$ around $1$, you get that there exist $x _n \in (1 - \frac 1 n , 1)$ such that $\log (1 - \frac 1 n) = \log 1 + \frac {\log ^{(1)} 1} {1} (-\frac 1 n) + \frac {\log ^{(2)} x _n} {2} (- \frac 1 n)^2 = - \frac 1 n - \frac 1 {2 x_n ^2 n^2}$, therefore $- \log (1 - \frac 1 n) = \frac 1 n + \frac {\theta _n} {2 n^2}$ where $\theta _n = \frac 1 {x_n ^2}$.
Since $1 - \frac 1 n < x_n < 1$, then $(1 - \frac 1 n)^2 < x_n ^2 < 1$, so $\frac 1 {(1 - \frac 1 n)^2} > \theta _n > 1$. This shows that $\theta _n > 1$, as required.
For 2), the above inequality shows that $\theta _n$ converges (by the squeeze theorem), so taking $\lim \limits _{n \to \infty}$ in it you get $1 \ge \lim \limits _{n \to \infty} \theta _n \ge 1$, so the required limit is $1$.