Find an N such that the given convergent sequence is within ε of its limit

Consider the convergent sequence $\{ a_n \}$ with terms defined by

$a_n = (-1)^n \left( \frac{9}{10} \right)^n.$

Let $L = \lim_{n \to \infty} a_n$ . Find the value of $L$ and the value of $N$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$ for each of the following values of $\varepsilon$ :

$\varepsilon = 1$ ,
$\varepsilon = 0.1$ ,
$\varepsilon = 0.01$ ,
$\varepsilon = 0.001$ ,
$\varepsilon = 0.0001$ .

First, we know

$\lim_{n \to \infty} (-1)^n \left( \frac{9}{10}\right)^n = \lim_{n \to \infty} \left(-\frac{9}{10}\right)^n = 0$

since $\left|\frac{-9}{10}\right| = \frac{9}{10} < 1$ , so by (10.10) on page 380 of Apostol we know the limit is 0.
Then we have,

$\begin{align*} |a_n - L| < \varepsilon && \implies && \left| \left(-\frac{9}{10} \right)^n \right| &< \varepsilon \\[9pt] && \implies && \left( \frac{9}{10} \right)^n &< \varepsilon \\[9pt] && \implies && \log \left(\frac{9}{10}\right)^n &< \log \varepsilon \\[9pt] && \implies && n \log \frac{9}{10} &< \log \varepsilon \\[9pt] && \implies && n &> \frac{\log \varepsilon}{\log \frac{9}{10}}. \end{align*}$

We reversed the inequality sign in the final step since $\log \frac{9}{10} < 0$ since $\frac{9}{10} < 1$ . Thus, if $N > \frac{\log \varepsilon}{\log \frac{9}{10}}$ then for every $n \geq N$ we have $|a_n| < \varepsilon$ . We compute for the given values of $\varepsilon$ as follows: