Does the sequence $T_n=2^{-n}$ converge? $$T_n=2^{-n}$$
How can I tell if this converges? with previous questions I have just let $n = \infty$, however I'm unsure about this one.
 A: You can tell (intuitively) that this converges by writing down a couple of elements and trying to find a pattern:
$$\frac12, \frac14, \frac18,\frac1{16}\dots$$
You should immediatelly see that the elements are becoming very small very fast. So, your intuition should tell you that the limit should be $0$. But that's only a small part of the deal.

This is mathematics, after all. All your hunches and ideas and intuition are empty if you cannot prove it.
Now, there are many ways to prove that the limit is $0$. The most basic way is to use the $\epsilon$-$\delta$ definition, which in this case is very easy to use. In fact:

If it takes you more than one minute to figure out how the $\epsilon$-$\delta$ proof of this limit looks like, you need to practice more. An excellent way to practice is to write down this exact proof. It isn't hard once you get the hang of it, and if you want to do anything in math, you need practice.


Of course, there are other ways of proving it. You could, for example, find some other sequence $S_n$ such that $0<T_n<S_n$ and $$\lim_{n\to\infty}S_n=0.$$
Then, if you know enough about limits, you can conclude that the limit of $T_n$ is also $0$.
A: Hint: a monotone decreasing sequence bounded from below converges.
(Similarly a monotone increasing sequence boundes from above converges also)
Of course this is a non constructive approach, as you only get the existance of a limit, and not what the limit actually is. But in some cases, this is already sufficient.
A: I think it is worth proving that $1/n\rightarrow 0$ rigorously with an epsilon-n proof at least once. After that, you can compare sequences like the one in your question to this sequence, since it is clear that
$$
2^{-n}\leq 1/n
$$
Hint: use the Archimedean property.
