Prove that $S_n = 1 − 2^{-n}$ converges to $1$ as $n$ approaches infinity using formal definition of convergence Let $S_n=1-2^{-n}$. Prove that the sequence $S_n$ converges to $1$ as $n$ approaches infinity. 
I let $N = \lceil\log_2 \varepsilon\rceil + 1$. Then there exists $k\geq 1$ such that $\varepsilon \geq 1/2^k$. Then, set $N=k+1$. Let $n\geq N$. 
I am having trouble proving this though. any help would be greatly appreciated. thank you!
 A: By definition of convergence you must show that, for every $\varepsilon>0$, there exists $N>0$ such that, for every $n\geq N$, it holds that 
$$
|1-(1-2^{-n})|=|2^{-n}|=2^{-n}<\varepsilon. 
$$
Or, rearranging the inequality, 
$$
\frac{1}{\varepsilon}<2^n.
$$
So it is quite obvious that this holds for all $n$ large enough. If you want an expression for $N$ in terms in $\varepsilon$ just take $log_2$ on both sides and you have
$$
\log_2{1/\varepsilon}< n.
$$
So you can just take N to be any natural number greater than $log_2(1/\varepsilon)$ (close but not quite what you wrote).
If you know how work with logarithms you can derive that $\log_2(1/\varepsilon)=\log_2(\varepsilon^{-1})=-\log_2(\varepsilon)$, so you are left with 
$$
-\log_2{\varepsilon} < n.
$$
A: Expansion of $2^n=(1+1)^n$ by the binomial theorem shows that $2^n>n.$ Thus $2^{-n}<\frac {1}{n}$ so $$|1-(1-2^{-n})|=2^{-n}< \frac {1}{n}$$. You can take it from there with $\epsilon$ and $\delta .$
A: From the following limit properties:

$$\lim_{x \to \infty} \frac{1}{n^{x}} = 0, \forall n>1$$
$$\lim_{x \to \infty}  f(x) \pm g(x) = \lim_{x \to \infty}  f(x) \pm \lim_{x \to \infty}  g(x)$$
$$ \lim_{x \to \infty}  c = c $$

where c is a constant.
We can get the answer:
$$\lim_{x \to \infty}  1 - 2^{-n}$$
$$= \lim_{x \to \infty}  1 - \frac{1}{2^n}$$
$$= \lim_{x \to \infty}  1 - \lim_{x \to \infty} \frac{1}{2^n}$$
$$=1 - 0 =1$$
