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  1. For $n \geq 2$, show that $\exists$ a number $\theta_n, \theta_n > 1$ such that $-\log(1-\frac{1}{n}) = \frac{1}{n} + \frac{\theta_n}{2n^2}$

  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.

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    $\begingroup$ Are you sure the statement is right? We have $$-\log\left(1\color{red}{-}\frac{1}{n}\right)=\frac{1}{n}+\frac{1}{2n^2}+O(n^{-3}).$$ $\endgroup$ – Jack D'Aurizio Jul 6 '15 at 18:29
  • $\begingroup$ Sorry, I don't have the question in front of me. But, as far as I remember, there wasn't any $O(n^{-3})$. $\endgroup$ – Jellyfish Jul 6 '15 at 18:32
  • $\begingroup$ @Jellyfish, the $O(n^{-3})$ simply refers to the error terms in that Maclaurin expansion Jack stated. For larger values of $n$, those error terms tend to $0$. $\endgroup$ – Prasun Biswas Jul 6 '15 at 18:36
  • $\begingroup$ Sorry, but we haven't started Maclaurin expansion/series yet. $\endgroup$ – Jellyfish Jul 6 '15 at 18:40
  • $\begingroup$ The question is mistaken: the left-hand side is negative while the right-hand side is positive. Please check things before posting, don't make us waste time with wrong questions (it should also be of interest to you: mistaken questions get downvoted). @JackD'Aurizio could be right, there could be a $-$ instead of $+$ inside the $\log$ function. $\endgroup$ – Alex M. Jul 6 '15 at 18:46
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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 $1$.

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