Differential and Integral Calculus 2021

Parts 1.-3. Results follow directly from the definition and by dividing it up into separate cases of the different signs of \(x\) and \(y\)

Part 4. Here we divide the triangle inequality into different cases.
Case 1.

First let \(x,y \geq 0\). Then it follows that \[\begin{aligned}|x+y|=x+y=|x|+|y|\end{aligned}\] and the desired inequality is shown.

Next let \(x,y < 0\). Then: \[\begin{aligned}|x+y|=-(x+y)=(-x)+ (-y)=|x|+|y|\end{aligned}\]

Finally we consider the case \(x\geq 0\) and \(y<0\). Here we have two subcases:

• For \(x \geq -y\) we have \(x+y\geq 0\) and thus \(|x+y|=x+y\) from the definition of absolute value. Because \(y<0\) then \(y<-y\) and therefore also \(x+y < x-y\). Overall we have: \[\begin{aligned}|x+y| = x+y < x-y = |x|+|y|\end{aligned}\]

• For \(x < -y\) then \(x+y<0\). We have \(|x+y|=-(x+y)=-x-y\). Because \(x\geq0\), we have \(-x < x\) and thus \(-x-y\leq x-y\). Overall we have: \[\begin{aligned}|x+y| = -x-y \leq x-y = |x|+|y|\end{aligned}\]

The case \(x<0\) and \(y\geq0\) we prove it analogously to the case 3, in which \(x\) and \(y\) are exchanged.

\(\square\)

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 \(|a_n|<\varepsilon\) for every \(n\geq n_0\) and an arbitrary \(\varepsilon\in\mathbb{R}\). But then we have \(|b_n|\leq|a_n|<\varepsilon\); 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 \[\begin{aligned}|a_n|<\frac{\varepsilon}{|c|}\end{aligned}\] for all \(n\geq n_0\). Rearranging we get: \[\begin{aligned} |c|\cdot|a_n|=|c\cdot a_n|<\varepsilon\end{aligned}\]

Part 4.

Because \((a_n)_n\) is a zero sequence, by the definition we have \(|a_n|<\frac{\varepsilon}{2}\) for all \(n\geq n_0\). Analogously, for the zero sequence \((b_n)_n\) there is a \(m_0 \in \mathbb{N}\) with \(|b_n|<\frac{\varepsilon}{2}\) for all \(n\geq m_0\).

Then for all \(n > \max(n_0,m_0)\) it follows (using the triangle inequality) that: \[\begin{aligned}|a_n + b_n|\leq|a_n|+|b_n|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon\end{aligned}\]

\(\square\)

To show this, 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: \[\begin{aligned}\left|\frac{2n^2+1}{n^2+1}-2\right| =&\left|\frac{2n^2+1-2\cdot\left(n^2+1\right)}{n^2+1}\right| \\ =&\left|\frac{2n^2+1-2n^2-2}{n^2+1}\right| \\ =&\left|-\frac{1}{n^2+1}\right| \\ =&\left|\frac{1}{n^2+1}\right| \\ <&\frac{1}{n},\end{aligned}\] The last step uses a crude approximation.

We let \(\varepsilon > 0\) be an arbitrary constant. Then we 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 proved the note above. Therefore \(a=2\) 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 \(|a_n - a|<\varepsilon\) 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 \(|a_{n}-a|<\varepsilon\) for \(n\geq n_{0}\), therefore \(|a_{\varphi(n)}-a|<\varepsilon\) 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: \[\begin{aligned}|a_n - a| <\ \varepsilon_1 := &\ \textstyle \frac{\varepsilon}{2|\lambda|} \text{ for all }n\geq n_0\\ |b_n - b| <\ \varepsilon_2 := &\ \textstyle \frac{\varepsilon}{2|\mu|} \text{ for all }n\geq n_1\end{aligned}\]

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: \[\begin{aligned}|b_n|\cdot|a_n-a| + |a|\cdot|b_n - b| <&\ B \cdot \left(|a_n - a| + |b_n - b| \right).\end{aligned}\] For all \(n>n_0\) we have \(|a_n - a|<\frac{\varepsilon}{2\cdot B}\) and \(|b_n - b|<\frac{\varepsilon}{2\cdot B}\), and - putting everything together - the desired inequality it shown.

\(\square\)