VERKKOKIRJA

Kohdat 1.-3. Results follow directly from the definition and by dividing it up into separate cases of the different signs of $x$ and $y$

Kohta 4. Tämä kohta voidaan todistaa neliöön korottamalla tai tutkimalla kaikki eri vaihtoehdot kuten alla.
Tapaus 1.

Olkoot $x,y \geq 0$. Silloin $\begin{aligned}|x+y|=x+y=|x|+|y|\end{aligned}$ ja kaava pätee.

Olkoot seuraavaksi $x,y < 0$. Silloin $\begin{aligned}|x+y|=-(x+y)=(-x)+ (-y)=|x|+|y|.\end{aligned}$

Tutkitaan lopuksi tapaus $x\geq 0$ and $y$, joka jakaantuu kahteen alakohtaan:

• Jos $x \geq -y$, niin $x+y\geq 0$ ja siten $|x+y|=x+y$ määritelmän perusteella. Koska $y$, niin $y$ ja sen vuoksi $x+y < x-y$. Siis $\begin{aligned}|x+y| = x+y < x-y = |x|+|y|.\end{aligned}$

• Jos $x < -y$, niin $x+y$, ja tällöin $|x+y|=-(x+y)=-x-y$. Koska $x\geq0$, niin $-x < x$ ja siten $-x-y\leq x-y$. Siispä $\begin{aligned}|x+y| = -x-y \leq x-y = |x|+|y|.\end{aligned}$

Jos $x$ ja $y\geq0$, niin väite seuraa samalla periaatteella kuin tapauksessa 3, kun vaihdetaan keskenään $x$ ja $y$.

$\square$

Parts 1 and 2. If $(a_n)_n$ is a zero sequence, then according to the definition there is an index $n_0 \in \mathbb{N}$, such that $|a_n|$ for every $n\geq n_0$ and an arbitrary $\varepsilon\in\mathbb{R}$. But then we have $|b_n|\leq|a_n|$; this proves parts 1 and 2 are correct.

Part 3. If $c=0$, then the result is trivial. Let $c\neq0$ and choose $\varepsilon > 0$ such that $\begin{aligned}|a_n|$ for all $n\geq n_0$. Rearranging we get: $\begin{aligned} |c|\cdot|a_n|=|c\cdot a_n|$

Part 4.

Because $(a_n)_n$ is a zero sequence, by the definition we have $|a_n|$ for all $n\geq n_0$. Analogously, for the zero sequence $(b_n)_n$ there is a $m_0 \in \mathbb{N}$ with $|b_n|$ for all $n\geq m_0$.

Then for all $n > \max(n_0,m_0)$ it follows (using the triangle inequality) that: $\begin{aligned}|a_n + b_n|\leq|a_n|+|b_n|$

$\square$

For a rigorous proof, we show that for every $\varepsilon > 0$ there exists an index $n_0\in\mathbb{N}$, such that for every term $a_n$ with $n>n_0$ the following relationship holds: $\left| \frac{2n^2+1}{n^2+1} - 2\right| < \varepsilon.$

Firstly we estimate the inequality: $\begin{aligned}\left|\frac{2n^2+1}{n^2+1}-2\right| =&\left|\frac{2n^2+1-2\cdot\left(n^2+1\right)}{n^2+1}\right| \\ =&\left|\frac{2n^2+1-2n^2-2}{n^2+1}\right| \\ =&\left|-\frac{1}{n^2+1}\right| \\ =&\left|\frac{1}{n^2+1}\right| \\ $

Now, let $\varepsilon > 0$ be an arbitrary constant. We then choose the index $n_0\in\mathbb{N}$, such that $n_0 > \frac{1}{\varepsilon} \text{, or equivalently, } \frac{1}{n_0} < \varepsilon.$ Finally from the above inequality we have: $\left|\frac{2n^2+1}{n^2+1}-2\right| < \frac{1}{n} < \frac{1}{n_0} < \varepsilon,$ Thus we have proven the claim and so by definition $a=2$ is the limit of the sequence.

$\square$

Assume $a\ne b$; choose $\varepsilon\in\mathbb{R}$ with $\varepsilon:=\frac{1}{3}|a-b|.$ Then in particular $[a-\varepsilon,a+\varepsilon]\cap[b-\varepsilon,b+\varepsilon]=\emptyset.$

Because $(a_{n})_{n}$ converges to $a$, there is, according to the definition of convergence, a index $n_{0}\in\mathbb{N}$ with $|a_{n}-a|< \varepsilon$ for $n\geq n_{0}.$ Furthermore, because $(a_{n})_{n}$ converges to $b$ there is also a $\widetilde{n_{0}}\in\mathbb{N}$ with $|a_{n}-b|< \varepsilon$ for $n\geq\widetilde{n_{0}}.$ For $n\geq\max\{n_{0},\widetilde{n_{0}}\}$ we have: $\begin{aligned}\varepsilon\ = &\ \frac{1}{3}|a-b| \Rightarrow\\ 3\varepsilon\ = &\ |a-b|\\ = &\ |(a-a_{n})+(a_{n}-b)|\\ \leq &\ |a_{n}-a|+|a_{b}-b|\\ < &\ \varepsilon+\varepsilon=2\varepsilon,\end{aligned}$ Consequently we have obtained $3\varepsilon\leq2\varepsilon$, which is a contradiction as $\varepsilon>0$. Therefore the assumption must be wrong, so $a=b$.

$\square$

We assume that the sequence $(a_n)_n$ has the limit $a$. By the definition of convergence, we have that $|a_n - a|$ for all $\varepsilon \in \mathbb{R}$ and $n\geq n_0$. Choosing $\varepsilon = 1$ gives:
$\begin{aligned}|a_n|-|a|&\ \leq |a_n -a| \\ &\ < 1,\end{aligned}$ And therefore also $|a_n|\leq |a|+1$.

Thus for all $n\in \mathbb{N}$: $|a_n|\leq \max \left\{ |a_1|,|a_2|,\ldots,|a_{n_0}|,|a|+1 \right\}=:r$

$\square$

By the definition of a subsequence $\varphi(n)\geq n$. Because $a_{n}\rightarrow a$ it is implicated that $|a_{n}-a|$ for $n\geq n_{0}$, therefore $|a_{\varphi(n)}-a|$ for these indices $n$.

$\square$

Part 1. Let $\varepsilon > 0$. We must show, that for all $n \geq n_0$ it follows that: $|\lambda \cdot a_n + \mu \cdot b_n - \lambda \cdot a - \mu \cdot b| < \varepsilon.$ The left hand side we estimate using: $|\lambda (a_n-a)+\mu (b_n - b)| \leq |\lambda|\cdot|a_n-a|+|\mu|\cdot|b_n-b|.$

Because $(a_n)_n$ and $(b_n)_n$ converge, for each given $\varepsilon > 0$ it holds true that: $\begin{aligned}|a_n - a|$

Therefore $\begin{aligned}|\lambda|\cdot|a_n-a|+|\mu|\cdot|b_n-b| < &\ |\lambda|\varepsilon_1 + |\mu|\varepsilon_2 \\ = &\ \textstyle{ \frac{\varepsilon}{2} + \frac{\varepsilon}{2} } = \varepsilon\end{aligned}$ for all numbers $n \geq \max \{n_0,n_1\}$. Therefore the sequence $\left( \lambda \left( a_n - a \right) + \mu \left( b_n - b \right) \right)_n$ is a zero sequence and the desired inequality is shown.

Part 2. Let $\varepsilon > 0$. We have to show, that for all $n > n_0$ $|a_n b_n - a b| < \varepsilon.$ Furthermore an estimation of the left hand side follows: $\begin{aligned} |a_n b_n - a b| =&\ |a_n b_n - a b_n + a b_n - ab| \\ \leq &\ |b_n|\cdot|a_n-a| + |a|\cdot|b_n - b|.\end{aligned}$ We choose a number $B$, such that $|b_n| \lt b$ for all $n$ and $|a| \lt b$. Such a value of $B$ exists by the Theorem of convergent sequences being bounded. We can then use the estimation: $\begin{aligned}|b_n|\cdot|a_n-a| + |a|\cdot|b_n - b|$ For all $n>n_0$ we have $|a_n - a|$ and $|b_n - b|$, and - putting everything together - the desired inequality it shown.

$\square$