Exercises: Subsequences, Accumulation points, and Cauchy sequences – "Math for Non-Geeks"

verweis=Math for Non-Geeks

Exercises: Subsequences

Exercise (Subsequences of the e-sequence)

Calculate the limit $lim_{n \to \infty} {(1 + \frac{1}{n^{2}})}^{n^{2}}$
Calculate the limit $lim_{n \to \infty} {(1 + \frac{1}{2 n})}^{n}$
Calculate the limit $lim_{n \to \infty} {(1 + \frac{1}{n^{2}})}^{n}$ using 1)

Solution (Subsequences of the e-sequence)

Subtask 1: As we know that the sequence $(a_{n}) = ({(1 + \frac{1}{n})}^{n})$ has the limit $lim_{n \to \infty} a_{n} = e$ , for the subsequence $(a_{k^{2}})_{k \in ℕ}$ holds that: $lim_{k \to \infty} a_{k^{2}} = lim_{k \to \infty} {(1 + \frac{1}{k^{2}})}^{k^{2}} = lim_{n \to \infty} {(1 + \frac{1}{n^{2}})}^{n^{2}} = e$

Subtask 2: With ${(1 + \frac{1}{2 n})}^{n} = {(1 + \frac{1}{2 n})}^{2 n \cdot \frac{1}{2}} = {({(1 + \frac{1}{2 n})}^{2 n})}^{\frac{1}{2}} = \sqrt{{(1 + \frac{1}{2 n})}^{2 n}}$ and $(a_{2 n}) = ({(1 + \frac{1}{2 n})}^{2 n})$ being a subsequence of the sequence $(a_{n})$ from 1, we get $lim_{n \to \infty} a_{2 n} = lim_{n \to \infty} {(1 + \frac{1}{2 n})}^{2 n} = e$ . Using the computation rules for limits, we get $lim_{n \to \infty} \sqrt{{(1 + \frac{1}{2 n})}^{2 n}} = \sqrt{lim_{n \to \infty} {(1 + \frac{1}{2 n})}^{2 n}} = lim_{n \to \infty} \sqrt{e}$

Subtask 3: For $(b_{n}) = ({(1 + \frac{1}{n^{2}})}^{n})$ holds that $b_{n} = \sqrt[n]{a_{n^{2}}}$ with the sequence $(a_{n^{2}}) = ({(1 + \frac{1}{n^{2}})}^{n^{2}})$ from subtask 1.

As $lim_{n \to \infty} a_{n^{2}} = e$ , $(a_{n^{2}})$ is bounded from above, there exists an $M > 0$ (actually $M > 1$ ) with $a_{n^{2}} \leq M$ for all $n \in ℕ$ . Therefore, $\sqrt[n]{M}$ is an upper bound of $(b_{n})$ . We even get $1 \leq b_{n} = {(1 + \frac{1}{n^{2}})}^{n} \leq \sqrt[n]{M}$ As $lim_{n \to \infty} \sqrt[n]{M} = 1$ , using the squeeze theorem we get $lim_{k \to \infty} b_{n} = lim_{n \to \infty} {(1 + \frac{1}{n^{2}})}^{n} = 1$

Exercises: Accumulation points

Exercise (Accumulation points)

Determine all accumulation points of the sequences $(a_{n})_{n \in ℕ}, (b_{n})_{n \in ℕ}$ with $a_{n} = \frac{1}{2^{n}} + (- 1)^{n}$ and $b_{n} = {(1 + \frac{(- 1)^{n}}{n})}^{n}$ .

How to get to the proof? (Accumulation points)

Regarding $(a_{n})$ : As $(\frac{1}{2^{n}})$ converges to $0$ , all of its subsequences converge to $0$ . The sequence $((- 1)^{n})$ alternates between $1$ and $- 1$ . Hence, its accumulation points are $1$ and $- 1$ .
Regarding $(b_{n})$ : For even $n$ -s, $(b_{n})$ is a subsequence of the $e$ -sequence ${(1 + \frac{1}{n})}^{n}$ . For odd $n$ -s, $(b_{n})$ is a subsequence of the sequence ${(1 - \frac{1}{n})}^{n}$ , which converges to $\frac{1}{e}$ .

Solution (Accumulation points)

As $a_{2 k} = \frac{1}{2^{2 k}} + (- 1)^{2 k} = \frac{1}{4^{k}} + 1 \overset{k \to \infty}{\to} 0 + 1 = 1$ , one accumulation point of $(a_{n})$ is $1$ .

Furthermore, we have $a_{2 k - 1} = \frac{1}{2^{2 k - 1}} + (- 1)^{2 k - 1} = \frac{2}{4^{k - 1}} - 1 \overset{k \to \infty}{\to} 0 - 1 = - 1$ . Hence, another accumulation point of $(a_{n})$ is $- 1$ .

These are the only accumulation points of $(a_{n})$ because every other subsequence of $(a_{n})$ either diverges or converges to $1$ or $- 1$ .

We have $b_{2 k} = {(1 + \frac{1}{2 k})}^{2 k} \overset{k \to \infty}{\to} e$ , as ${(1 + \frac{1}{n})}^{n} \overset{n \to \infty}{\to} e$ . Thus, an accumulation point of $(b_{n})$ is $e$ .

Furthermore, we have $b_{2 k + 1} = {(1 - \frac{1}{2 k + 1})}^{2 k + 1} \overset{k \to \infty}{\to} \frac{1}{e}$ , as ${(1 - \frac{1}{n})}^{n} \overset{n \to \infty}{\to} \frac{1}{e}$ . Therefore, another accumulation point of $(b_{n})$ is $\frac{1}{e}$ .

These are the only accumulation points of $(b_{n})$ because every other subsequence of $(b_{n})$ either diverges or converges to $e$ or $\frac{1}{e}$ .

Exercises: Lim inf and Lim sup

Exercise (Lim inf and Lim sup)

Determine Lim inf and Lim sup of the sequences $(a_{n})_{n \in ℕ}, (b_{n})_{n \in ℕ}$ with $a_{n} = {(\frac{3}{2} + (- 1)^{n})}^{n}$ and $b_{n} = {\begin{matrix} 2 + \frac{1}{3^{n}} & for n = 3 k, \\ 3 + \frac{n + 2}{n} & for n = 3 k + 1, \\ 3 & for n = 3 k + 2 \end{matrix}$ .

How to get to the proof? (Lim inf and Lim sup)

We start off investigating whether $(a_{n})$ is bounded from above/below. This sequence is bounded from below, but not from above. Hence, Lim inf equals the smallest accumulation point of $(a_{n})$ . We can determine it looking at the subsequence, including the odd sequence elements. As the sequence is not bounded from above, Lim sup equals $\infty$ .
As all of its subsequences are bounded, $(b_{n})$ is bounded from above and below. In order to determine Lim sup and Lim inf of $(b_{n})$ , we need to find the limits of its subsequences.

Solution (Lim inf and Lim sup)

As $\frac{3}{2} + (- 1)^{n} \geq 0$ , we have $a_{n} \geq 0$ . Hence, $(a_{n})$ is bounded from below. $lim inf a_{n}$ then equals the smallest accumulation point of the sequence $(a_{n})$ . For the subsequence $(a_{2 k - 1})$ holds that

$lim_{k \to \infty} a_{2 k - 1} = lim_{k \to \infty} {(\frac{3}{2} + (- 1)^{2 k - 1})}^{2 k - 1} = lim_{k \to \infty} {(\frac{3}{2} - 1)}^{2 k - 1} = lim_{k \to \infty} {(\frac{1}{2})}^{2 k - 1} = 0$ Therefore, $0$ is an accumulation point of $(a_{n})$ . There can't be other accumulation points, as all other subsequences of our sequence either converge to $0$ or diverge. Hence, we have $\underset{n \to \infty}{lim inf} a_{n} = 0$ .

Alternative solution: We have $({\tilde{b}}_{k})_{k \in ℕ} = (inf {a_{n} : n \geq k})_{k \in ℕ} = (0, 0, 0, 0, \dots)$ , as the odd sequence elements of $(a_{n})_{n \in ℕ}$ monotonically decrease to $0$ . It follows that $\underset{n \to \infty}{lim inf} a_{n} = lim_{k \to \infty} b_{k} = 0$

With $n = 2 k$ , we get $a_{2 k} = {(\frac{3}{2} + (- 1)^{2 k})}^{2 k} = {(\frac{3}{2} + 1)}^{2 k} = {(\frac{5}{2})}^{2 k} = {(\frac{25}{4})}^{k}$ The Archimedean property then states that there exists a $k \in ℕ$ with $a_{2 k} = {(\frac{25}{4})}^{k} > S$ for all $S > 0$ . Therefore, $(a_{n})$ is not bounded from above and $\underset{n \to \infty}{lim sup} a_{n} = \infty$ .

Regarding $(b_{n})$ : We have

$\begin{matrix} b_{3 k} & = 2 + \frac{1}{3^{3 k}} = 2 + \frac{1}{2 7^{k}} \overset{k \to \infty}{\to} 2 + 0 = 2 \\ b_{3 k + 1} & = 3 + \frac{3 k + 3}{3 k + 1} = 3 + 1 + \frac{2}{3 k + 1} \overset{k \to \infty}{\to} 4 + 0 = 4, \\ b_{3 k + 2} & = 3 \overset{k \to \infty}{\to} 3. \end{matrix}$

Hence, $(b_{n})$ only has the accumulation points $2, 3$ and $4$ . Furthermore, $(b_{n})$ is bounded, as all subsequences are bounded as well. We get $\underset{n \to \infty}{lim inf} b_{n} = 2$ and $\underset{n \to \infty}{lim sup} b_{n} = 4$ .

Exercise (Lim inf and Lim sup 2)

Let $(a_{n})_{n \in ℕ}$ be a sequence of positive real numbers. Prove the following inequalities: $\underset{n \to \infty}{lim inf} \frac{a_{n + 1}}{a_{n}} \leq \underset{n \to \infty}{lim inf} \sqrt[n]{a_{n}} \leq \underset{n \to \infty}{lim sup} \sqrt[n]{a_{n}} \leq \underset{n \to \infty}{lim sup} \frac{a_{n + 1}}{a_{n}}$
Let $(a_{n})_{n \in ℕ}$ be a sequence with $a_{n} = {\begin{matrix} \frac{1}{2^{n}} & for n = 2 k - 1, \\ \frac{1}{3^{n}} & for n = 2 k \end{matrix}$ Compute $\underset{n \to \infty}{lim inf} \frac{a_{n + 1}}{a_{n}}$ , $\underset{n \to \infty}{lim inf} \sqrt[n]{a_{n}}$ , $\underset{n \to \infty}{lim sup} \sqrt[n]{a_{n}}$ and $\underset{n \to \infty}{lim sup} \frac{a_{n + 1}}{a_{n}}$ .
Derive the following statement from 1): Let $(a_{n})_{n \in ℕ}$ be a sequence of positive real numbers with $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = a \in ℝ$ . Then we have $lim_{n \to \infty} \sqrt[n]{a_{n}} = a$ .
Prove the following limits: $lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}} = 0 and lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}} = e$

Solution (Lim inf and Lim sup 2)

Let $(a_{n})$ be a positive sequence with
$α = lim inf \frac{a_{n + 1}}{a_{n}}, α^{'} = lim inf \sqrt[n]{a_{n}}, β^{'} = lim sup \sqrt[n]{a_{n}}, β = lim sup \frac{a_{n + 1}}{a_{n}}$
We need to show the following inequalities:
$0 \leq α \leq α^{'} \leq β^{'} \leq β \leq \infty$
W.l.o.g. let $α > 0$ and $β < \infty$ . By definition, $α$ is the smallest and $β$ the largest accumulation point of the sequence ${(\frac{a_{n + 1}}{a_{n}})}_{n \in ℕ}$ . Hence, for all $ε > 0$ ( $< α$ ) there exists an $m \in ℕ$ with
$α - ε < \frac{a_{k + 1}}{a_{k}} < β + ε \forall k \geq m$
Applying the inequality $(n - m)$ times setting $k = m, \dots, n - 1$ , for an arbitrary $n \in ℕ$ with $n - 1 \geq m$ holds that:
$(α - ε)^{n - m} < \frac{a_{m + 1}}{a_{m}} \cdot \frac{a_{m + 2}}{a_{m + 1}} \cdot \dots \cdot \frac{a_{n - 1}}{a_{n - 2}} \cdot \frac{a_{n}}{a_{n - 1}} < (β + ε)^{n - m}$
The expression in the middle is a telescoping product. Except for the numerator $a_{n}$ and the denominator $a_{m}$ , everything cancels out.
$(α - ε)^{n - m} < \frac{a_{n}}{a_{m}} < (β + ε)^{n - m}$
Multiplying $a_{m}$ and taking the $n$ -th root leads to
$\sqrt[n]{a_{m}} \cdot (α - ε)^{1 - \frac{m}{n}} < \sqrt[n]{a_{n}} < \sqrt[n]{a_{m}} (β + ε)^{1 - \frac{m}{n}}$
With $lim_{n \to \infty} \sqrt[n]{a_{m}} = 1$ , the term on the left side converges to $α - ε$ and the one on the right side converges to $β + ε$ for $n \to \infty$ . Therefore, we get
$α - ε \leq lim inf \sqrt[n]{a_{n}}$
As $ε$ can be chosen arbitrarily small, we get
$α \leq α^{'}$
As the inequality $α^{'} \leq β^{'}$ holds by definition, the claim follows.
- For all $n \in ℕ$ we have $\frac{a_{n + 1}}{a_{n}} \geq 0$ , as the numerator and denominator are always positive. Hence, 0 is a lower bound of the quotient sequence $(\frac{a_{n + 1}}{a_{n}})$ . With $n = 2 k - 1$ being odd and $n + 1 = 2 k$ even, it follows that
$\frac{a_{n + 1}}{a_{n}} = \frac{a_{2 k}}{a_{2 k - 1}} = \frac{\frac{1}{3^{2 k}}}{\frac{1}{2^{2 k - 1}}} = \frac{2^{2 k - 1}}{3^{2 k}} = \frac{2^{2 k}}{2 \cdot 3^{2 k}} = \frac{1}{2} \cdot \frac{4^{k}}{9^{k}} = \frac{1}{2} \cdot {(\frac{4}{9})}^{k} \to 0$ for $k \to \infty$ Hence, there exists a subsequence of the quotient sequence converging to zero. It follows that lim infn→∞an+1an=0
- With $n = 2 k$ being even and $n + 1 = 2 k + 1$ being odd, we get
$\frac{a_{n + 1}}{a_{n}} = \frac{a_{2 k + 1}}{a_{2 k}} = \frac{\frac{1}{2^{2 k + 1}}}{\frac{1}{3^{2 k}}} = \frac{3^{2 k}}{2^{2 k + 1}} = \frac{3^{2 k}}{2 \cdot 2^{2 k}} = \frac{1}{2} \cdot \frac{9^{k}}{4^{k}} = \frac{1}{2} \cdot {(\frac{9}{4})}^{k} \to \infty$ for $k \to \infty$ Hence, the quotient series doesn't have an upper bound and lim supn→∞an+1an=∞
- With $n = 2 k - 1$ being odd, we have
$\sqrt[n]{a_{n}} = \sqrt[2 k - 1]{a_{2 k - 1}} = \sqrt[2 k - 1]{\frac{1}{2^{2 k - 1}}} = \frac{1}{2}$ and n=2k being even leads to $\sqrt[n]{a_{n}} = \sqrt[2 k]{a_{2 k}} = \sqrt[2 k]{\frac{1}{3^{2 k}}} = \frac{1}{3}$ Hence, (ann) has the accumulation points 12 and 13. It follows that $\underset{n \to \infty}{lim inf} \sqrt[n]{a_{n}} = \frac{1}{3}$ and $\underset{n \to \infty}{lim sup} \sqrt[n]{a_{n}} = \frac{1}{2}$
As $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = a$ , we get $\underset{n \to \infty}{lim inf} \frac{a_{n + 1}}{a_{n}} = \underset{n \to \infty}{lim sup} \frac{a_{n + 1}}{a_{n}} = a$ Using 1) and the squeeze theorem, we have $\underset{n \to \infty}{lim inf} \sqrt[n]{a_{n}} = \underset{n \to \infty}{lim sup} \sqrt[n]{a_{n}} = a$ It follows that $lim_{n \to \infty} \sqrt[n]{a_{n}} = a$ .
1st limit: Let $a_{n} = \frac{1}{n!}$ . It follows that
$\begin{matrix} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} & = lim_{n \to \infty} \frac{\frac{1}{(n + 1)!}}{\frac{1}{n!}} \\ = lim_{n \to \infty} \frac{n!}{(n + 1)!} \\ = lim_{n \to \infty} \frac{n!}{(n + 1) n!} \\ = lim_{n \to \infty} \frac{1}{n + 1} = 0 \end{matrix}$
Using part 3, we have $lim_{n \to \infty} \sqrt[n]{a_{n}} = lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}} = lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}} = 0$ .
2nd limit: Let $a_{n} = \frac{n^{n}}{n!}$ . It follows that
$\begin{matrix} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} & = lim_{n \to \infty} \frac{\frac{(n + 1)^{n + 1}}{(n + 1)!}}{\frac{n^{n}}{n!}} \\ = lim_{n \to \infty} \frac{(n + 1)^{n + 1} n!}{n^{n} (n + 1)!} \\ = lim_{n \to \infty} \frac{(n + 1)^{n} (n + 1) n!}{n^{n} (n + 1) n!} \\ = lim_{n \to \infty} \frac{(n + 1)^{n}}{n^{n}} \\ = lim_{n \to \infty} {(\frac{n + 1}{n})}^{n} \\ = lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e \end{matrix}$
Again using 3. , we get $lim_{n \to \infty} \sqrt[n]{a_{n}} = lim_{n \to \infty} \sqrt[n]{\frac{n^{n}}{n!}} = lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}} = e$ .

Hint

Take a look at the inequalities proven in subtask 1. In subtask 2 even holds that $\underset{n \to \infty}{lim inf} \frac{a_{n + 1}}{a_{n}} < \underset{n \to \infty}{lim inf} \sqrt[n]{a_{n}} < \underset{n \to \infty}{lim sup} \sqrt[n]{a_{n}} < \underset{n \to \infty}{lim sup} \frac{a_{n + 1}}{a_{n}}$

Hint

Taking a look at the limits from subtask 4 and using reciprocals, we get: $lim_{n \to \infty} \sqrt[n]{n!} = \infty and lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$

Exercises: Cauchy sequences

Exercise (Cauchy criterion for sequences)

Let $0 < q < 1$ and $(a_{n})_{n \in ℕ}$ be a sequence of real numbers with $| a_{n + 1} - a_{n} | \leq q \cdot | a_{n} - a_{n - 1} |$ for all $n \in ℕ$ . Show that the sequence $(a_{n})_{n \in ℕ}$ converges.
Let $a_{0} = 0, a_{1} = 1$ and $a_{n + 1} = \frac{1}{2} (a_{n} + a_{n - 1})$ . Show that the sequence $(a_{n})_{n \in ℕ}$ converges and compute its limit.

Solution (Cauchy criterion for sequences)

Subtask 1: We will show that $(a_{n})$ is a Cauchy sequence. The Cauchy criterion then states that $(a_{n})$ converges. Using the assumptions made, we have: $| a_{j + 1} - a_{j} | \leq q \cdot | a_{j} - a_{j - 1} | \leq q^{2} \cdot | a_{j - 1} - a_{j - 2} | \leq \dots \leq q^{j} \cdot | a_{1} - a_{0} | .$ If $m, n \in ℕ$ and $m > n$ , there exists a $k \in ℕ$ with $m = n + k$ . It follows that

$\begin{matrix} | a_{m} - a_{n} | & = | a_{n + k} - a_{n} | \\ = | a_{n + k} - a_{n + k - 1} + a_{n + k - 1} - a_{n + k - 2} + \dots + a_{n + 2} - a_{n + 1} + a_{n + 1} - a_{n} | \\ ↓ generalized triangle inequality \\ \leq | a_{n + k} - a_{n + k - 1} | + | a_{n + k - 1} - a_{n + k - 2} | + \dots + | a_{n + 2} - a_{n + 1} | + | a_{n + 1} - a_{n} | \\ = \sum_{j = 0}^{k - 1} | a_{n + j + 1} - a_{n + j} | \\ ↓ index shift \\ = \sum_{j = n}^{n + k - 1} | a_{j} - a_{j - 1} | \end{matrix}$

From the inequalities follows that

$\begin{matrix} | a_{m} - a_{n} | & \leq \sum_{j = n}^{n + k - 1} | a_{j} - a_{j - 1} | \\ \leq \sum_{j = n}^{n + k - 1} q^{j} \cdot | a_{1} - a_{0} | \\ = | a_{1} - a_{0} | q^{n} \sum_{j = 0}^{k - 1} q^{j} \\ ↓ geometric sum formula \\ = | a_{1} - a_{0} | q^{n} \frac{1 - q^{k}}{1 - q} \\ ↓ q^{k} \geq 0 \Rightarrow 1 - q^{k} \leq 1 \\ \leq \frac{| a_{1} - a_{0} |}{1 - q} q^{n} \end{matrix}$

From the Archimedean axiom we can derive that for all $ϵ > 0$ there exists an $N \in ℕ$ such that for all $n, m \geq N$ holds $| a_{m} - a_{n} | \leq \frac{| a_{1} - a_{0} |}{1 - q} q^{n} < ϵ$ Hence, $(a_{n})$ is a Cauchy sequence and therefore convergent.

Subtask 2: For the sequence $(a_{n})$ holds true that $| a_{n + 1} - a_{n} | = | \frac{1}{2} (a_{n} + a_{n - 1}) - a_{n} | = | \frac{1}{2} a_{n - 1} - \frac{1}{2} a_{n} | = \frac{1}{2} | a_{n} - a_{n - 1} | .$ With $q = \frac{1}{2}$ , the criterion from subtask 1 is met. Hence, $(a_{n})$ converges.

The limit can be computed as follows: Analogous to subtask 1 - but without taking the absolute values -, we have $a_{n + 1} - a_{n} = \frac{1}{2} (a_{n} + a_{n - 1}) - a_{n} = \frac{1}{2} (a_{n - 1} - a_{n}) = - \frac{1}{2} (a_{n} - a_{n - 1}) .$ Applying the formula $n$ times leads to $a_{n + 1} - a_{n} = {(- \frac{1}{2})}^{n} (a_{1} - a_{0}) .$ Hence, we get $a_{n + 1} - \underset{= 0}{\underset{⏟}{a_{0}}} \underset{sum}{\overset{telescoping}{=}} \sum_{k = 0}^{n} (a_{k + 1} - a_{k}) \underset{above}{\overset{see}{=}} \sum_{k = 0}^{n} {(- \frac{1}{2})}^{k} \underset{= 1 - 0 = 1}{\underset{⏟}{(a_{1} - a_{0})}} \underset{sum formula}{\overset{geometric}{=}} \frac{1 - {(- \frac{1}{2})}^{n + 1}}{1 + \frac{1}{2}} = \frac{1 - {(- \frac{1}{2})}^{n + 1}}{\frac{3}{2}} = \frac{2}{3} \cdot (1 - {(- \frac{1}{2})}^{n + 1}) .$ Using the computation rules for limits, we have $lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} a_{n} = lim_{n \to \infty} \frac{2}{3} \cdot (1 - {(- \frac{1}{2})}^{n + 1}) = \frac{2}{3} \cdot (1 - lim_{n \to \infty} {(- \frac{1}{2})}^{n + 1}) = \frac{2}{3} \cdot (1 - 0) = \frac{2}{3}$

Series →

Feedback? Do you want to join?

If you have questions concerning the content, or didn't understand something, the feel free to contact us! We would love to answer your questions! Also we are thankful for critics and/or comments! If you share our vision to explain university math in an comprehensible way, then contact us under:

E-Mail: en@serlo.org

E-Mail: hochschulmathematik@serlo.org

This article is licensed under the free license CC-BY-SA 3.0. With that you can use it, modify it or share it freely, as long as you name „Serlo“ as source and put you changes under the same CC-BY-SA 3.0 oder an compatible license. On the page „Kopier uns!“ we explain you what you have to pay attention to, when using our texts, picture or videos.