I'm trying to understand how I prove the convergence of a sequence and I think I am getting somewhere slowly after watching a bunch of videos and reading up about it.
Now I used the following sequence to practice a bit and I am unsure if I am done with it.
$\prod_{k=2}^{n} {(1-\frac{1}{k^2})}$ which can be written in $a_n = \frac{n+1}{2n}$. The limit as $n \rightarrow \infty$ is $\frac{1}{2}$.
Now to the proof, according to the definition $\forall \epsilon > 0 \exists n_0 \in \mathbb{N}:|a_n-a|<\epsilon \forall n \geq n_0$
$|\frac{n+1}{2n}-\frac{1}{2}| = |\frac{1}{2n}|= \frac{1}{2n} \leq \frac{1}{2n_0} < \epsilon$
therefore
$\frac{1}{2n_0} \leq \frac{1}{n_0} < \epsilon$
$\Rightarrow \frac{1}{\epsilon} < n_0$
Now after the definition $n \geq n_0$ I have to find something bigger than $\frac{1}{\epsilon}$
$\Rightarrow$ $\frac{2}{\epsilon} = n_0$ and since its in $\mathbb{N}$ I just round up with the gaussian brackets $\lceil \frac{2}{\epsilon} \rceil = n_0$. Now I have my $n_0$ and can write out the whole proof.
Let $\epsilon > 0$ be arbitary. Pick $n_0 := \lceil \frac{2}{\epsilon} \rceil$. Than $|a_n-a| = |\frac{n+1}{2n}-\frac{1}{2}| = |\frac{1}{2n}|= \frac{1}{2n} \leq \frac{1}{2n_0} \leq \frac{1}{n_0} = \frac{1}{\lceil \frac{2}{\epsilon} \rceil} \leq \frac{1}{2\epsilon} < \epsilon$
q.e.d
Thats what I have done now and I am not sure if everything is correct.. if so I guess I finally understood the conecept of it. I hope someone can take a look at it and tell me where I might have messed up.
Greetings
The underlying idea makes sense. I would probably make the structure of the argument stand out a little more:
Pick an $\epsilon > 0$. We must then find an $n_0$ such that $|\frac{n+1}{2n} - \frac{1}{2}| < \epsilon$ for all $n \geq n_0$.
We have $|\frac{n+1}{2n} - \frac{1}{2}| = |\frac{1}{2n}| = \frac{1}{2n}$ (as $n$ is positive).
To ensure that $\frac{1}{2n} < \epsilon$ we must have $n > \frac{1}{2 \epsilon}$. So take $n_0 = \lceil \frac{1}{2 \epsilon} \rceil$.