Winter 2016
MATH 226: Limits and infinite series
Branko Ćurgus
- Friday, February 12, 2016
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- Today in class I handed out the
new assignment which is due next Friday.
- Thursday, February 11, 2016
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- Theorem. Let $a : \mathbb N \to \mathbb R$ and $b : \mathbb N \to \mathbb R$ be two sequences and $K, L \in \mathbb R$.
Assume:
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$\displaystyle \lim_{n\to+\infty} a_n = K$.
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$\displaystyle \lim_{n\to+\infty} b_n = L$.
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There exists $n_0 \in \mathbb N$ such that for all $n \in \mathbb N$ such that $n \geq n_0$ we have $a_n \leq b_n$.
Then $K \leq L$.
- Proof. Assume that the conditions 1, 2 and 3 in the theorem are satisfied.
Let $\epsilon \gt 0$ be arbitrary.
By definition of convergence for sequences the condition 1. implies that there exists $N_a(\epsilon) \in \mathbb R$ such that
\[
n \in \mathbb N \quad \text{and} \quad n \gt N_a(\epsilon) \quad \Rightarrow \quad |a_n - K | \lt \epsilon.
\]
Notice that the condition $ |a_n - K | \lt \epsilon$ is equivalent to $K - \epsilon \lt a_n \lt K + \epsilon$. Therefore the last displayed implication can be rewritten as
\[ \tag{G1}
n \in \mathbb N \quad \text{and} \quad n \gt N_a(\epsilon) \quad \Rightarrow \quad K - \epsilon \lt a_n \lt K + \epsilon.
\]
Similarly, by definition of convergence for sequences the condition 2. implies that there exists $N_b(\epsilon) \in \mathbb R$ such that
\[ \tag{G2}
n \in \mathbb N \quad \text{and} \quad n \gt N_b(\epsilon) \quad \Rightarrow \quad L - \epsilon \lt b_n \lt L + \epsilon.
\]
Let $m \in \mathbb N$ be such that $m \gt \max\bigl\{n_0, N_a(\epsilon), N_b(\epsilon)\bigr\}$.
From (G1), the condition 3. and (G2) we deduce that
\begin{align*}
K- \epsilon \lt & \ a_m \lt K+\epsilon \\
& \ a_m \leq b_m \\
L & - \epsilon \lt b_m \lt L+\epsilon.
\end{align*}
From the last three displayed relations we deduce that
\[
K- \epsilon \lt a_m \leq b_m \lt L+\epsilon.
\]
Consequently,
\[
K - L \lt 2 \epsilon.
\]
Now recall that $\epsilon \gt 0$ was arbitrary. Since the inequality $K - L \lt 2 \epsilon$ holds for all $\epsilon \gt 0$ we conclude that $K - L \leq 0$. Hence $K \leq L$ and this completes the proof.
- Friday, January 22, 2016
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In this file I summarize steps involved in limit proofs for
 = L.)
.
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Do limits in Exercises 4.10.
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Pay special attention to limits of specific functions in which the real number $a$ is not specified. For example, in class we proved
\[
\forall \ a \gt 0 \qquad \lim_{x\to a} \frac{1}{x} = \frac{1}{a} \qquad \text{and} \qquad \forall \ a \in \mathbb R \qquad \lim_{x\to a} x^2 = a^2.
\]
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Do Exercise 4.11, Exercise 4.12 and limits in Exercises 4.15.
- Monday, January 11, 2016
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- In
this file I summarize steps involved in limit proofs for
 = L.)
.
- As a very simple example how to use Mathematica
here is the file that I created today.
- Another example of a Mathematica notebook is
this Mathematica file. In this file I show how to explore functions in Mathematica. The file is called PlottingFunctions.nb. Right-click on the underlined word "Here"; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name.
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After saving the file you can open it with Mathematica. For this file use Mathematica 5.2. (We also have Mathematica 8 in BH 215. These two versions are not compatible.) You will find Mathematica 5.2 on most Windows computers on campus; click here for a list of labs with Mathematica installed (select Mathematica from the long list of programs and click search). To locate Mathematica on a particular computer you might try
Start -> All Programs -> Math Applications -> Mathematica.
Open Mathematica first; then open PlottingFunctions.nb from Mathematica. You can execute the entire file by the following manu sequence (in Mathematica):
Kernel -> Evaluation -> Evaluate Notebook.
There are some more instructions in the file.
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To get started with Mathematica 5.2 see my
Mathematica page.
- Tuesday, January 5, 2016
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