# Exercises: sequences – Serlo

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• Introduction • Complex numbers • Supremum and infimum • Sequences • Sequences • Explicit and recursive description • Examples and properties of sequences • Exercises • Convergence and divergence • Subsequences, Accumulation points and Cauchy sequences • Series • Convergence criteria for series • Exponential and Logarithm functions • Trigonometric and Hyperbolic functions • Continuity • Differential Calculus ## Questions: Sequences

Exercise

Which of the following statements are true?

1. Every index $n$ can uniquely be matched to a sequence element $a_{n}$ .
2. Every sequence element $a_{n}$ can uniquely be matched to an index $n$ .
3. A sequence being neither bounded from above nor from below does not exist.
4. A sequence increasing and decreasing monotonically does not exist.
5. A sequence increasing and decreasing strictly monotonically does not exist.
6. A sequence increasing monotonically and being bounded from above does not exist.

Solution

1. True.
2. False. A constant sequence is a great counterexample. As all sequence elements $a_{n}$ , $n\in \mathbb {N}$ , have the same value, you can match this value to every index $n\in \mathbb {N}$ and not only one unique index.
3. False. The sequence $\left(a_{n}\right)_{n\in \mathbb {N} }=1,\,-2,\,3,\,-4,\,5,\,-6,\,\ldots$ is neither bounded from above nor from below.
4. False. Every constant sequence is monotonically increasing and decreasing. Only „strict monotonicity“ implies that the sequence elements need to be unequal.
5. True.
6. False. The sequence $(a_{n})_{n\in \mathbb {N} }$ with $a_{n}:=1-{\tfrac {1}{n}}$ for all $n\in \mathbb {N}$ is bounded from above, as $a_{n}<1$ holds for all $n\in \mathbb {N}$ . Furthermore, it's monotonically increasing. For all $n\in \mathbb {N}$ , we have ${\tfrac {1}{n+1}}<{\tfrac {1}{n}}$ and therefore $a_{n}=1-{\tfrac {1}{n}}<1-{\tfrac {1}{n+1}}=a_{n+1}$ .

## Exercise: Find a sequence

Exercise

Find a sequence $(a_{n})_{n\in \mathbb {N} }$ fulfilling the following conditions:

1. $a_{n} for all $n\in \mathbb {N}$ 2. $a_{n}>a_{n+1}$ for all odd $n$ Write down an explicit and a recursive formula for your sequence!

Solution

One possible sequence is

$(a_{n})_{n\in \mathbb {N} }=1,0,2,1,3,2,4,3,5,4,6,5,7,6,\ldots$ An explicit formula of this sequence is given by

$a_{n}={\begin{cases}{\frac {n+1}{2}}&{\text{for }}n{\text{ odd}}\\{\frac {n}{2}}-1&{\text{for }}n{\text{ even}}\\\end{cases}}$ A recursive formula of this sequence is given by $a_{1}:=1,a_{2}:=0$ and for all $n\geq 2$ :

$a_{n}:=a_{n-2}+1$ 