Proof that the sequence (0.9. 0.99. 0.999, . . . ) converges to 1. Is it correct? I was attempting to prove that the limit of the sequence defined by $(\frac{10^n - 1}{10^n})^{\infty}_{n=1}$ converges to 1. Would really appreciate if someone could confirm if the proof is correct or not. Here is the proof.
Proof: Let $\varepsilon > 0$ be a real number. Note that for $n \geq N$ we have that $|a_n -1|\leq |a_N-1|=|\frac{1}{10^n}|\leq|\frac{1}{10^N}|$. We want a choice for $N$ such that $\frac{1}{10^N}<\varepsilon$. This will ensure that for all $n\geq N$, $|a_n-1|<\varepsilon$. Manipulating the inequality $\frac{1}{10^N}< \varepsilon$
gives $10^N > \frac{1}{\varepsilon}$. We take log base $10$ on both sides. $log_{10}(10^N) > log_{10}(1) - log_{10}(\varepsilon)$. From which we get that $N > -log_{10}(\varepsilon)$.
Thus, for any $\epsilon >0$ pick $N>-log_{10}(\varepsilon)$. Then for all $n\geq N$ we have that $|a_n -1|<\varepsilon$.
Hence, $\lim_{n \to \infty} (\frac{10^n - 1}{10^n}) = 1$
Is this correct?

This bit isn't correct:

$|a_n -1|\leq |a_N-1|=|\frac{1}{10^n}|\leq|\frac{1}{10^N}|$

Most people can guess that you mean
$|a_n -1|=|\frac{1}{10^n}|\leq|\frac{1}{10^N}|= |a_N-1|$

We take log base $10$ on both sides. $log_{10}(10^N) > log_{10}(1) - log_{10}(\varepsilon)$.

Here I think you need to mention that $\log$ is strictly increasing, since you have actually used that fact when you said the inequality holds after taking logarithms.
Otherwise it looks basically OK.
