Show: $\log(1-\frac{1}{n}) + \epsilon > - \frac{1}{n}$ for all but finitely many $n$

Question

I am trying to show that, for $\epsilon > 0$, we have $\log(1-\frac{1}{n}) + \epsilon > - \frac{1}{n}$ for all but finitely many $n$. I was thinking about using Borel-Cantelli lemma but I don't know where I should start. I would appreciate any suggestions.

Answer 1

There is a power series, $$\log(1-x) = -\Big( x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots \Big) $$ which converges for all $-1 < x < 1$. It can be derived by integrating the geometric series $1/(1-x) = 1+x+x^2+\cdots$.

Then for $0 < x < 1$, $$ 0 < \frac{x^2}{2} + \frac{x^3}{3} + \cdots < \frac{x^2}{2}\Big( 1 + x + x^2 + \cdots\Big) = \frac{x^2}{2(1-x)}.$$ It should now be plain that the right hand side converges to zero as $x \to 0$ and we can find some $x_0$ that ensures it is less than $\varepsilon$ for all $x < x_0$. Accordingly for all $n > 1/x_0$ the same holds with $x$ replaced by $1/n$. For such $n$, it follows $$ -\log\Big(1-\frac{1}{n}\Big) - \frac{1}{n} < \varepsilon $$ and noting that all but finaitely many $n$ will exceed $1/x_0$, you have your result.