How do I show that a sequence is convergent.

Question

I feel like theres a huge hole in my understanding of sequences and convergence.

I want to show that the sequence $(\frac{n}{2^n})_{n\in\mathbb{N}}$ is convergent and find its limit. It's obvious that the limit of this sequence is $0$, but I'm just not sure how to prove that this sequence is convergent

I know the definition of convergence is that for every $\epsilon > 0$ there exists an $N_{\epsilon}$ such that $|\frac{n}{2^n} - l| < \epsilon$ for $n>N_{\epsilon}$. This is as far as I've gotten with my attempt.

$\textbf{Proof:}$

$\lim_{n \to \infty} \frac{n}{2^n} = 0$

For $\epsilon > 0, \>\exists \>N_{\epsilon} \in \mathbb{N}$ s.t

$|\frac{n}{2^n} - 0| < \epsilon$ for all $n > N_{\epsilon}$

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

???

I just can't seem to find an $N_{\epsilon}$ that guarantees that $\frac{n}{2^n} < \epsilon$ for $n > N_{\epsilon}$

Am I on the right track? If not, what am I missing? I seem to struggle with proving the convergence of any sequence. Is there some sort of step by step system that I can follow to determine whether or not a sequence is convergent? Any help or hints would be greatly appreciated!

Answer 1

Guide:

When $n$ is large, $2^n > n^2$.

That is when $n$ is large, we have $\frac1{2^n} <\frac1{n^2} $

That is when $n$ is large, we have $\frac{n}{2^n} < \frac{n}{n^2}=\frac1n$.

$|\frac{n}{2^n}| < |\frac{n}{n^2}|=|\frac1n|$.

If you can prove that $\frac1n$ converges to $0$, then you can show that $\frac{n}{2^n}$ converges to $0$.

Answer 2

Here is an interesting solution that is not direct.

If $n>1$, $\frac{n+1}{2^{n+1}}=\frac{\frac{n+1}{2}}{2^n}<\frac{n}{2^n}$. Also, $\frac{1}{2^1} = \frac{2}{2^2}$. So the sequence is decreasing. Clearly, it is bounded below by $0$. So it is convergent (say to $L$).

Say $x_n = \frac{n}{2^n}$. Then $L = \lim x_{n+1} = \lim \frac{n+1}{2n}\cdot x_n=\frac{L}{2}$. So $L=0$

Answer 3

$(2)^n= = (1+1)^n = $

$1+n +n(n-1)/(2!) +.....+1\gt$

$ n(n-1)/(2!).$

$|\dfrac{n}{2^n}|\lt |\dfrac{2n}{(n-1)(n)}|= |\dfrac{2}{n-1}|.$

Let $\epsilon >0 .$

Archimedes:

There exists a $n_0 -1 \in \mathbb{Z^+}$ such that $n_0-1 > 2/\epsilon.$

For $n > n_0$ we have .

$|\dfrac{n}{2^n}|\lt |\dfrac{2}{n-1}| \lt |\dfrac{2}{n_0-1} |\lt \epsilon. $