# Convergence

Convergent sequences have a well-defined limit.

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Concept | Content |
---|---|

Logical Statements and Operations | Logic is the foundation to formulate proofs and to understand the language of mathematics. |

Real Numbers | In a real analysis, the real numbers are the largest number set we need. They satisfy axioms that represent the idea of a number line. |

Sequences | These object are needed to define limits later on. |

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Concept | Content |
---|---|

Bounded Sequences | Sequences can be bounded from above and from below. The values of a bounded sequence cannot become arbitrarily large. |

Limit Inferior and Limit Superior | The largest and smallest limit of all convergent subsequences. |

Limit Theorems | Combining limits is a useful tool to deduce convergence and the limit of a more complicated sequence from the convergence of simpler building blocks. |

Monotonicity of Limits and Sandwich Theorem | Taking limits respects weak inequalities and sandwiching a sequence by two converging sequences reveals its limit. The technique of sandwiching can be used to determine limits via nested intervals. |

Cauchy Sequences | The sequence members of a Cauchy Sequence eventually become arbitrarily close to each other. |

Open, Closed, Compact Sets | Important notions for subsets of real numbers. |

Subsequences and Accumulation Values | A sequence that does not converge may still have converging subsequences. |

Bolzano-Weierstrass Theorem | Every bounded sequence has at least one converging subsequence. |

Series and Partial Sums | A series is a sequence of partial sums. |

Limits of Functions | How function evaluations change when the argument approaches a certain point. |

Continuity | The concept that relates functions with convergent sequences. |

Bounded Monotonic Sequences | If a sequence of real numbers is bounded and monotonic, then it is convergent. |

## Study *Convergence* #

Next we define the notions of convergence and limits:

**Definition 1** (Convergence/divergence of sequences).
*Let \((a_n)_{n\in\mathbb{N}}\) be
a sequence in \(\mathbb{K}\). We say
that*

*\((a_n)_{n\in\mathbb{N}}\) is**convergent to \(a\in \mathbb{K}\)*if for all \(\varepsilon>0\) there exists some \(N=N(\varepsilon)\in\mathbb{N}\) such that for all \(n\geq N\) holds \(|a_n-a|<\varepsilon\). In this case, we write \[\lim_{n\to\infty}a_n=a.\]*\((a_n)_{n\in\mathbb{N}}\) is**divergent*if it is not convergent, i.e., for all \(a\in \mathbb{K}\) holds: There exists some \(\varepsilon>0\) such that for all \(N\) there exists some \(n>N\) with \(|a_n-a|\geq\varepsilon\).

Convergence for real sequences means that if you give any small
distance \(\varepsilon\), one finds
that all sequence members \(a_n\) lie
in the interval \((a-\varepsilon,
a+\varepsilon)\) with the exception of only *finitely*
many.

**Example 2**.

Show: \((a_{n})_{n \in \mathbb{N}}\) with \(a_n = (1/n)\) is convergent with limit \(0\).

Show: \((b_{n})_{n \in \mathbb{N}}\) with \(b_n = (1/\sqrt{n})\) is convergent with limit \(0\).

*Proof.*Let \(\varepsilon > 0\). Choose \(N > \frac{1}{\varepsilon^2}\). Then for all \(n \geq N\), we have \[|b_n - 0| = \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{N}} < \varepsilon\] This means \(b_n\) is arbitrarily close to \(0\), eventually. ◻Show: \((d_{n})_{n \in \mathbb{N}}\) with \(\displaystyle d_n = \frac{1}{\sqrt{1 + \frac{1}{n}}}\) is convergent with limit \(1\).

*Proof.*We know that \(\sqrt{1 + 1/n}\) is always greater than \(1\). Therefore, we can calculate: \[|1 - \frac{1}{\sqrt{1 + \frac{1}{n}}} |^2 \leq \frac{ 1- 2 \sqrt{1 + \frac{1}{n}} + 1 + \frac{1}{n}}{1 + \frac{1}{n}}\] \[\leq \frac{ 1- 2 + 1 + \frac{1}{n}}{1 + \frac{1}{n}} = \frac{ \frac{1}{n}}{1 + \frac{1}{n}} \leq \frac{1}{n}\] The square root is a monotonically increasing function, and hence, we can put the square root to both sides and get the inequality: \[0 ~\leq~ 1 - \frac{1}{\sqrt{1 + \frac{1}{n}}} ~\leq~ \frac{1}{\sqrt{n}}\] By the sandwich theorem of Homework 1, this shows that the sequence \((e_n)_{n \in \mathbb{N}N}\) given by \(e_n := 1 - \frac{1}{\sqrt{1 + \frac{1}{n}}}\) is convergent with limit \(0\). Then we use the limit theorem to get: \[\lim_{n \rightarrow \infty} d_n = \lim_{n \rightarrow \infty} (1 - e_n) = \lim_{n \rightarrow \infty} 1 - \lim_{n \rightarrow \infty} e_n = 1 - 0 = 1\] ◻

**Remark 3**.

It can be shown that for a complex sequence \((a_n)_{n\in\mathbb{N}}\), convergence to \(a\in\mathbb{C}\) holds true if, and only if, \((\Re(a_n))_{n\in\mathbb{N}}\) converges to \(\Re(a)\)

__and__\((\Im(a_n))_{n\in\mathbb{N}}\) converges to \(\Im(a)\)In fact, convergence can also be defined for sequences in some arbitrary normed vector space \(V\) (for the definition of a normed space, e.g. consult the linear algebra script). Then one has to replace the absolute value by the norm (e.g., “\(\|a_n-a\|<\varepsilon\)”.)

Due to the fact that for any \(N\in\mathbb{N}\), we can find some \(x\in\mathbb{R}\) with \(x>N\), we can equivalently reformulate the convergence definition as follows: “\((a_n)_{n\in\mathbb{N}}\) is convergent to \(a\in \mathbb{R}\) if for all \(\varepsilon>0\) there exists some \(N=N(\varepsilon)\in\mathbb{R}\) such that for all \(n\geq N\) holds \(|a_n-a|<\varepsilon\)”. In the following, we just write “there exists some \(N\)”.

What does convergence mean?

Outside any \(\varepsilon\)-neighbourhood of \(a\) only finitely many elements of the sequence exist.

**Example 4**.

The real sequence \((\frac1n)_{n\in\mathbb{N}}\) converges to 0.

*Proof:*Let \(\varepsilon>0\) (be arbitrary): Choose \(N=\frac1\varepsilon+1=\frac{1+\varepsilon}\varepsilon\).

Then for all \(n\geq N\) holds \[\frac1n\leq\frac1N=\frac{\varepsilon}{1+\varepsilon}<\varepsilon.\] Therefore \[\left|\frac1n-0\right|=\frac1n<\varepsilon.\]The real sequence \(((-1)^n)_{n\in\mathbb{N}}\) is divergent.

*Proof by contradiction:*

Assume that \(((-1)^n)_{n\in\mathbb{N}}\) is convergent to \(a\in\mathbb{R}\). Then take e.g. \(\varepsilon=\frac1{10}\). Due to convergence, we should have some \(N\) such that for all \(n\geq N\) holds \[|(-1)^n-a|<{\textstyle\frac1{10}}.\] Thus we have that \(|-1-a|<\frac1{10}\) and \(|1-a|<\frac1{10}\). As a consequence, \[2=|1+a-a+1|\leq|1+a|+|-a+1|=|1+a|+|a-1|<{\textstyle\frac1{10}+\frac1{10}=\frac1{5}}.\] This is a contradiction.\(\Box\)For \(q\in\mathbb{C}\backslash\{0\}\) with \(|q|<1\) the complex sequence \((q^n)_{n\in\mathbb{N}}\) converges to 0.

*Proof:*\(|q|<1\) gives rise to \(\frac1{|q|}>1\), whence \(\frac1{|q|}-1>0\). Therefore, we are able to apply Bernoulli’s inequality (see tutorial) in the following way:\[\frac1{|q|^n}=\left(1+\left(\frac1{|q|}-1\right)\right)^n =\left(1+\left(\frac{1-|q|}{|q|}\right)\right)^n\geq 1+n\cdot \left(\frac{1-|q|}{|q|}\right),\] and thus \[|q|^n\leq \frac1{1+n\cdot \left(\frac{1-|q|}{|q|}\right)}=\frac{|q|}{|q|+n\cdot ({1-|q|})}.\] Now let \(\varepsilon>0\) (be arbitrary):

Choose \[N=\frac{|q|}{\varepsilon\cdot(1-|q|)}-\frac{|q|}{1-|q|}+1\] Then for all \(n\geq N\) holds \[n>\frac{|q|}{\varepsilon\cdot(1-|q|)}-\frac{|q|}{1-|q|}\] and thus \[n\cdot ({1-|q|})>\frac{|q|}{\varepsilon}-|q|.\] This leads to \[|q|+n\cdot ({1-|q|})>\frac{|q|}{\varepsilon},\] whence \[\frac{|q|}{|q|+n\cdot ({1-|q|})}<\varepsilon.\] The above calculations now imply \[|q^n-0|=|q|^n\leq\frac{|q|}{|q|+n\cdot ({1-|q|})}<\varepsilon.\] \(\Box\)

**Remark 5**. The choice of the \(N\) often seems “to appear from nowhere”.
However, there is a systematic way to formulate the proof. For instance
in a), we need to end up with the equation \(|\frac1n-0|<\varepsilon\) or,
equivalently, \(\frac1n<\varepsilon\). Inverting this
expression leads to \(n>\frac1\varepsilon\). Therefore, if
\(N\) is chosen as \(N=\frac1\varepsilon+1=\frac{1+\varepsilon}\varepsilon\),
the desired statement follows.

If one has to formulate such a proof (for instance, in some exercise), then first these above calculations have to be done “on some extra sheet” and then formulate the convergence proof in the style as in a) or c).

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