Whether or not $\{ 2^{-n}\}$ converges

Question

So I was trying to prove this theorem using the epsilon definition of the limit, but am currently a little stuck!

So far what I have is this:

From the definition of the convergence of a sequence to a limit we hav4 that, a sequence is said to converge to some limit l, if $\forall \epsilon > 0, \exists M, st:$ if $ n \ge M, |a_{n} - l|< \epsilon$. So we show that when l = 0:

$|\frac{1}{2^{n}}-0| < \epsilon$. So we find such $\epsilon $ and $M$ that satisfies the definition above. And we say:

$|\frac{1}{2^{n}}-0| < \epsilon$

$\implies \frac{1}{2^{n}} < \epsilon$

But what do I do from here? Natural logs have not been defined, neither has it been defined what it means to raise something to the ($\frac{1}{n}$). Can someone please help?

Answer 1

$2^{n}=(1+1)^{n} > n$ by Binomial Theorem. So $\frac 1 {2^{n}} <\frac 1 n <\epsilon$ if $n >\frac 1 {\epsilon}$.

Answer 2

In your definition, you forget to consider (and write) that $M$ is a function of $\varepsilon$, that is, $M=M(\varepsilon)$, that is, as consequence, $\varepsilon=f_M=\frac{1}{2^M}$. Now you know that for $n>M$

$$ |a_{n} - l|< a_{M}\,(=\varepsilon), $$

which is the case, because for $n>M$

$$ \frac{1}{2^n}<\frac{1}{2^M}. $$

Which I think should be your answer.