Show that, for $n\in \mathbb{N}$, the sequence is converging. Problem: Show that, for $n\in \mathbb{N}$, the sequence
$$a_n=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2^2}\right)\cdot...\cdot\left(1-\frac{1}{2^n}\right),$$
is convergent.

The sequence should converge if the general factor (last factor) approaches 1. So we can form the function
$$f(x) = 1-\frac{1}{2^x},$$
only by looking at this, it's abundantly clear that $1/2^x\rightarrow 0$ as $x\rightarrow \infty.$ This means that 
$$\lim_{x\rightarrow\infty}f(x)=1,$$
and convergence is shown.
This seems a bit too easy to be enough of a proof. Does this suffice or am I missing something critical?
 A: I would say that it is that easy, but your approach isn't the right one. Rather, this is a monotonously decreasing sequence bounded by $0$, so it must converge. Whether that limit is $0$ or something else requires some more careful calculation, though.
To see why your approach is flawed in general, consider what would happen if there were a $+$ inside the brackets instead of a $-$. In that case, you would have a product of terms that come closer and closer to $1$, but with more and more terms. There is no easy way to conclude whether that converges or diverges. For instance, $$b_n=\left(1+\frac11  \right)\left(1+\frac12  \right)\cdots\left(1+\frac1n  \right)=n+1$$ diverges.
A: In order to determine if $(a_n)_{n\in\mathbb N}$ converges to $0$ or not, you can check if $\lim_{n\to\infty}\log a_n=-\infty$ or not. Note that$$\log a_n=\log\left(1-\frac12\right)+\log\left(1-\frac1{2^2}\right)+\cdots+\log\left(1-\frac1{2^n}\right).$$So, we are dealing with a series here. The terms are negative, but it converges to $-\infty$ if and only if the series$$\sum_{n=1}^\infty-\log\left(1-\frac1{2^n}\right)\tag{1}$$converges to $+\infty$. But$$\lim_{x\to0}\frac{-\log(1-x)}x=1$$and therefore the divergence of the series $(1)$ is equivalent to the divergence of the series $\sum_{n=1}^\infty\frac1{2^n}$. But this series converges. Therefore, $\lim_{n\to\infty}a_n>0$.
A: Correct me if wrong.
$a_n \gt 0$, $ n \in \mathbb{Z+}.$
$a_{n+1} = a_n (1-\dfrac{1}{2^{n+1}}) \lt a_n.$
Hence $a_n$, positve, is strictly monotonically decreasing , bounded below.
$\rightarrow :$
Convergent.
A: First, the sequence $a_n$ is monotone 
$$
a_1>a_2>a_3>\cdots >a_n>\cdots
$$
Second, the sequence $a_n$ is limited. For $0<x<1$ we have $ln(1-x)<x$. Then 
$$
\sum_{k=1}^{n}\ln\left(1-\frac{1}{2^k}\right)
\leq
\sum_{k=1}^{n}\frac{1}{2^k}
$$
and
$$
|a_n|=|e^{\log a_n}|
=
\left|\;e^{\sum_{k=1}^{n}\ln\left(1-\frac{1}{2^k}\right)}\;\right|
\leq
\left|e^{\sum_{k=1}^{n}\frac{1}{2^k}}\right|
=
e^{\frac{1/2-(1/2)^n}{1-1/2}}
\leq 
e^{\frac{1/2}{1-1/2}}=e
$$
