I'm studying calculus, and recently I've found a task I cannot solve: prove Weierstrass approximation theorem for a sequence and find its limit. The sequence is:
0.2, 0.23, 0.233, ...
I described it like this:
$x_n = x_{n-1} + 3 * 10^{-n}, n => \infty$ (idk how to format the n -> inf properly)
It is easily proven that each next element is bigger than previous by $3 * 10^{-(n+1)}$ (because $x_n + 3 * 10^{-(n+1)} > x_n$), which means it is monotonous. But I have no idea how to prove that the sequence is limited by something ($\exists\lim{x_n}$).
Moreover, according to the way we're taught: lim(X) = lim(function describing this X) => X = {function describing this X}
which leads me to these steps:
1) $\lim{x_n} = \lim{x_{n-1} + 3 * 10^{-n}}$
2) $x = x + 3 * 10^{-n}$
3) $0 = 3 * 10^{-n}$
How do I prove the existence of the $\lim{x_n}$, and how do I find it, so it makes sense, contrary to the $0 = 3 * 10^{-n}$ ?
$x_1=0.2$
$x_n=x_{n-1}+3\cdot 10^{-n},n>1$
Means that the sequence is
$\{0.2,0.23,0.233,0.2333,\ldots\}$
that is $$\lim_{n\to\infty}x_n=0.2+3 \sum _{n=2}^{\infty } \left(\dfrac{1}{10}\right)^n=0.2+\dfrac{1}{30}=\dfrac{7}{30}=0.2\bar 3$$
and this is the limit because the geometric series with ratio $\dfrac{1}{10}$ converges