So I have a sequence $\{a_n\} = \frac{\pi}{2^n}$ where $n=1,2,3,4 \dots$ And I need to prove that its limit is $0$. Here is what have done, can someone check and tell me if this is correct?
Definition: A sequence $\{a_n\}$ has a limit $L$ if and only if for any $\epsilon>0$, there exists $N \in \mathbb{N}$ such that $\forall n>N,|a_n-L|<\epsilon$.
A sequence that has a limit $L$ is called a convergent sequence and we say that the sequence converges to $L$. If a sequence $\{a_n\}$ converges to a limit $L$, we write $\{a_n\} \to L$ as $n \to \infty$ or $\lim_{n \to \infty}$ of $\{a_n\}=L$. If a sequence does not have a limit, then we say that the sequence diverges.
So for our sequence given that $\epsilon>0$ we want $|\frac{\pi}{2^n} |<\epsilon~~ \forall n>N$.
So $|\frac{\pi}{2^n} |<\epsilon =\frac{\pi}{2^n} <\epsilon$ Because $n$ is strictly positive making the whole left side of the inequality positive.
Solving for $n$:
$$\frac{\pi}{2^n} <\epsilon$$
$$2^n<\frac{\pi}{\epsilon}$$
$$n \log 2 < \log \frac{\pi}{\epsilon}$$
$$n < \log( \frac{\pi}{\epsilon}-2)$$
So if you take N to be any integer such that $N> \log(\frac{\pi}{\epsilon}-2)$, then for all $\epsilon >0$ we have $|\frac{\pi}{2^n} |<\epsilon~~ \forall n>N$, and hence we have shown that $\lim_{n \to \infty}$ of $\{A_n\}=L$
Sorry that it looks like this but I don't know how to format text in here.
You have the right idea. Once you get to the point $\frac{\pi}{2^n} < \epsilon$, the algebra gives $n>\frac{\log\left(\pi/\epsilon\right)}{\log{2}}$. Your solution switched the order of the inequality, and brought the 2 into the log incorrectly.