Proposition 2.2.1.
Assume that \(X\to L\) and \(c\in\mathbb{R}\text{.}\) Define \(cX=(cy_n)_{n\in\mathbb{N}}\text{.}\) Then \(cX\to cL\text{.}\)
Section 2.2 algebra and convergence
We already saw one compatibility between convergence and algebra, in the form of Proposition 2.1.4. Here are some others, which we’ll prove in class. They’re recorded here just to make sure we’ve got them all in one place.
Proposition 2.2.2.
Assume that \(X\to L\) and \(Y\to M\text{.}\) Define \(XY=(x_ny_n)_{n\in\mathbb{N}}\text{.}\) Then \(XY\to LM\text{.}\)
Proposition 2.2.3.
Assume that \(X\to L\) and \(Y\to M\text{.}\) Further, assume that \(y_n\neq 0\) for all \(n\text{,}\) and \(M\neq 0\text{.}\) Define \(\frac{X}{Y}=\left(\frac{x_n}{y_n}\right)_{n\in\mathbb{N}}\text{.}\) Then \(\frac{X}{Y}\to \frac{L}{M}\text{.}\)
Definition 2.2.4.
The \(K\)-tail of the sequence \(X=(x_n)_{n\in\mathbb{N}}\) is the sequence
\begin{equation*}
X_{(K)}=(x_{n+K})_{n\in\mathbb{N}}
\end{equation*}
Checkpoint 2.2.5.
Proposition 2.2.6.
The following are equivalent:
-
\(\displaystyle X\to L\)
-
There is a \(K\in\mathbb{N}\) so that \(X_{(K)}\to L\text{.}\)
-
For any \(K\in\mathbb{N}\text{,}\) \(X_{(K)}\to L\)
