nyuuzyou/wikis huggingface dataset:
Convergence of P-Series/Lemma

== Lemma for Convergence of P-Series ==
Let $p = x + i y$ be a complex number where $x, y \in \R$ such that:
:$x > 0$
:$x \ne 1$
Then:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges $\ds \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ converges.
== Proof ==
Let $p = x + i y$.
Then:
by Euler's Formula.
Now since $x > 0$, and all $n \ge 1$, all terms are positive and we may do away with the absolute values.
Then by the Cauchy Integral Test:
:$\ds \sum_{n \mathop = 1}^{\to \infty} \frac 1 {n^x}$ converges $\ds \int_1^\infty \frac {\d t} {t^x}$ converges.
First let $x \ne 1$:
Hence:
:$\ds \int_1^{\to \infty} \frac {\d t} {t^x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} } - \dfrac {1^{1 - x} } {1 - x}$
:
:$\ds \int_1^{\to \infty} \frac {\d t} {t^x} + \dfrac {1^{1 - x} } {1 - x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} }$
and so:
:$\ds \sum_{n \mathop = 1}^\infty \size {n^{-p} }$ converges
:
:$\ds \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ converges.
For $x = 1$:
which diverges.
== Sources ==
* : $\S 6.6$
* : $\S 13.34 \ (3)$
* : $\S 1.2.2$: Summary of convergence tests
Category:Convergence of P-Series
Convergence of P-Series/Real/Proof 1

== Theorem ==
== Proof ==
By the Cauchy Integral Test:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges the improper integral $\ds \int_1^\infty \frac {\d t} {t^x}$ exists.
The result follows from Integral to Infinity of Reciprocal of Power of x.
== Sources ==
* : $\S 13.34 \ (3)$
* : $\S 1.2.2$: Summary of convergence tests
Category:Convergence of P-Series
Convergence of P-Series/Real/Proof 2

== Theorem ==
== Proof ==
Let $p = 1$.
Then from Harmonic Series is Divergent the $p$-series diverges.
So let $p > 1$.
We note that the sequence of partial sums is increasing.
Hence it is sufficient to show that they are bounded above.
Let:
:$s_{2^N} := 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dotsb + \dfrac 1 {N^p}$
Then:
Hence the result.
== Sources ==
* : $\S 6.6$
Category:Convergence of P-Series
Convergence of Product of Convergent Scalar Sequence and Convergent Vector Sequence in Normed Vector Space

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {V, \norm \cdot}$ be a normed vector space on $\Bbb F$.
Let $\alpha \in \R$.
Let $x \in V$.
Let $\sequence {\alpha_n}_{n \mathop \in \N}$ be a real sequence in $\Bbb F$ such that: 
:$\alpha_n \to \alpha$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $V$ such that:
:$x_n \to x$
in $\struct {V, \norm \cdot}$. 
Then we have: 
:$\alpha_n x_n \to \alpha x$
in $\struct {V, \norm \cdot}$.
== Proof ==
Let $\epsilon > 0$.
We have: 
From Convergent Real Sequence is Bounded, there exists $M > 0$ such that we have: 
:$\size {\alpha_n} \le M$
for each $n \in \N$. 
Then we have: 
:$\norm {\alpha_n x_n - \alpha x} \le M \norm {x_n - x} + \norm x \size {\alpha_n - \alpha}$
Suppose that: 
:$\norm x = 0$
then, from positive definiteness of $\norm \cdot$: 
:$x = 0$ 
In this case:
:$x_n \to 0$
and:
:$\norm {\alpha_n x_n} = \norm {\alpha_n x_n - \alpha x} \le M \norm {x_n - x} = M \norm {x_n}$
Then we can choose $N \in \N$ such that: 
:$\ds \norm {x_n} < \frac \epsilon M$
for $n > N$.
From which we have: 
:$\norm {\alpha_n x_n} < M$ 
for $n > N$. 
So, from the definition of a convergent sequence, we have: 
:$\alpha_n x_n \to 0 = \alpha x$
in the case $\norm x = 0$.
Now suppose that: 
:$\norm x \ne 0$
Now suppose that: 
:$\norm x \ne 0$
Since: 
:$\alpha_n \to \alpha$
we can pick $N_1 \in \N$ such that: 
:$\ds \size {\alpha_n - \alpha} < \frac \epsilon {2 \norm x}$
for $n \ge N_1$, from the definition of convergence.
Since we also have: 
:$x_n \to x$
we can pick $N_2 \in \N$ such that: 
:$\ds \norm {x_n - x} < \frac \epsilon {2 M}$
from the definition of a convergent sequence.
Let:
:$N = \max \set {N_1, N_2}$
We then have, for $n > N$: 
So, from the definition of a convergent sequence, we have: 
:$\alpha_n x_n \to \alpha x$
in the case $x \ne 0$ also, hence the result.
Category:Convergent Sequences (Normed Vector Spaces)
Category:Normed Vector Spaces
Convergence of Sequence in Discrete Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $S$.
Then $\sequence {x_n}_{n \mathop \in \N}$ converges in $T$ to a limit :
:$\exists N \in \N: \forall n \in \N: n > N \implies x_n = x_N$
That is, the sequence reaches some value of $S$ and "stays there".
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
Suppose $\sequence {x_n}_{n \mathop \in \N}$ converges to a limit $L$.
Then:
=== Necessary Condition ===
Let $N \in \N$ such that:
:$\forall n \in \N: n > N \implies x_n = x_N$
Let $U$ be an open neighborhood of $x_N$.
Then:
:$\forall n \in \N: n > N \implies x_n = x_N \in U$
Thus $\sequence {x_n}_{n \mathop \in \N}$ converges to $x_N$.
Category:Convergent Sequences
Category:Discrete Topology
Convergence of Sequence in Discrete Space/Corollary

== Corollary to Convergence of Sequence in Discrete Space ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $S$.
Let $\sequence {x'_n}_{n \mathop \in \N}$ be a sequence of distinct terms in $S$.
Then $\sequence {x'_n}_{n \mathop \in \N}$ is not convergent in $T$.
== Proof ==
By the definition of a sequence of distinct terms:
:$\forall x \in \sequence {x'_n}_{n \mathop \in \N}: r \ne s \implies x_r \ne x_s$
Hence trivially:
:$\neg \exists k \in \N: \forall m \in \N: m > k: x'_m = x'_k$
The result follows from Convergence of Sequence in Discrete Space.
Category:Convergent Sequences
Category:Discrete Topology
Convergence of Sequence in Subspace

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $T_H = \struct{H, \tau_H}$ be a topological subspace of $T$, where $H \subseteq S$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $S$.
Then $\sequence {x_n}_{n \mathop \in \N}$ converges to $p$ in $T_H$ $\sequence {x_n}_{n \mathop \in \N}$ converges to $p$ in $T$.
== Proof ==
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces
Category:Convergent Sequences
Category:Topological Subspaces
Convergence of Sequence of Test Functions in Test Function Space implies Convergence in Schwartz Space

== Theorem ==
Let $\map \DD \R$ be the test function space.
Let $\map \SS \R$ be the Schwartz space.
Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence of test functions in $\map \DD \R$.
Let $\mathbf 0 : \R \to 0$ be the zero mapping.
Suppose $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$:
:$\phi_n \stackrel \DD \longrightarrow \mathbf 0$
Then $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \SS \R$:
:$\phi_n \stackrel \SS \longrightarrow \mathbf 0$
== Proof ==
For all $n \in \N$ let $\phi_n$ be a test function.
By definition, $\phi_n$ has a compact support $I_n \subset \R$:
:$\forall x \notin I_n \implies \map {\phi_n} x = 0$
Let:
:$a \in \R : a > 0 : \forall n \in \N : I_n \subseteq \closedint {-a} a$
Then:
Suppose:
:$\phi_n \stackrel \DD \longrightarrow \mathbf 0$
Then all derivatives of $\phi_n$ on $\closedint {-a} a$ converge uniformly to $\mathbf 0$.
Hence:
In other words:
:$\ds \forall m,k \in \N : \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^k \map {\phi^{\paren m}_n} x} = 0$
By definition:
:$\phi_n \stackrel \SS \longrightarrow \mathbf 0$
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Test Functions
Category:Uniform Convergence
Category:Convergence
Convergence of Series of Complex Numbers by Real and Imaginary Part

== Theorem ==
Let $\sequence {z_n}$ be a sequence of complex numbers.
Then:
:the series $\ds \sum_{n \mathop = 1}^\infty z_n$ converges to $Z \in \C$
:
:the series:
::$\ds \sum_{n \mathop = 1}^\infty \map \Re {z_n}$
:and:
::$\ds \sum_{n \mathop = 1}^\infty \map \Im {z_n}$
:converge to $\map \Re Z$ and $\map \Im Z$ respectively.
== Proof ==
Let:
:the $n$th partial sum of $\sequence {z_n}$ be denoted $Z_n$
:the $n$th partial sum of $\sequence {\map \Re {z_n} }$ be denoted $U_n$
:the $n$th partial sum of $\sequence {\map \Im {z_n} }$ be denoted $V_n$
Then:
:$Z_n = U_n + i V_n$
Let:
:$\lim_{n \mathop \to \infty} U_n = U$
:$\lim_{n \mathop \to \infty} V_n = V$
By definition of convergent complex sequence:
:$\lim_{n \mathop \to \infty} Z_n = \lim_{n \mathop \to \infty} U_n + i \lim_{n \mathop \to \infty} V_n$
and so:
:$\lim_{n \mathop \to \infty} Z_n = U + i V$
and the result follows.
== Sources ==
* : $\S 4.3$. Series
Category:Convergent Complex Sequences
Category:Complex Series
Convergence of Square of Linear Combination of Sequences whose Squares Converge

== Theorem ==
Let $\sequence {x_i}$ and $\sequence {y_i}$ be real sequences such that the series $\ds \sum_{i \mathop \ge 0} {x_i}^2$ and $\ds \sum_{i \mathop \ge 0} {y_i}^2$ are convergent.
Let $\lambda, \mu \in \R$ be real numbers.
Then $\ds \sum_{i \mathop \ge 0} \paren {\lambda x_i + \mu y_i}^2$ is convergent.
== Proof ==
Let $n \in \N$.
Then:
:$\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2 = \lambda^2 \sum_{i \mathop = 1}^n {x_i}^2 + \mu^2 \sum_{i \mathop = 1}^n {y_i}^2 + 2 \lambda \mu \sum_{i \mathop = 1}^n x_i y_i$
By Cauchy's Inequality:
:$\ds \sum_{i \mathop = 1}^n x_i y_i \le \paren {\sum_{i \mathop = 1}^n {x_i}^2}^{\frac 1 2} \paren {\sum_{i \mathop = 1}^n {y_i}^2}^{\frac 1 2}$
Hence:
Thus the sequence of partial sums $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is bounded above.
We also have that $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is also increasing.
So by the Monotone Convergence Theorem, $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is convergent.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.16$
Category:Real Analysis
Category:Series
Convergence of Taylor Series of Function Analytic on Disk/Lemma/Proof 1

== Lemma ==
== Proof ==
Let $z$ be a real number.
We have:
:$\ds \sum_{n \mathop = 0}^\infty n z^n$ has radius of convergence of $1$
by Complex Power Series/Examples/n.
This leads to that:
:$\ds \sum_{n \mathop = 0}^\infty n \paren {\frac 1 y}^n$ converges
as $\ds \frac 1 y < 1$ as $y > 1$.
This leads to that:
:the sequence $\ds \sequence {\frac n {y^n}}$ converges to the limit $0$
by Terms in Convergent Series Converge to Zero.
This means that:
:$\ds \lim_{n \mathop \to \infty} \frac n {y^n}$ = $0$
by definition of limit.
This concludes the proof.
Category:Real Analysis
Convergent Complex Sequence/Examples

== Examples of Convergent Complex Sequence ==
=== Example: $\dfrac {\paren {3 + i n}^2} {n^2}$ ===
=== Example: $\paren {\dfrac {1 + i n} {1 + n} }^3$ ===
=== Example: $\paren {\cos \dfrac \pi n + i \sin \dfrac \pi n}^{2 n + 1}$ ===
=== Example: $\paren {\dfrac 1 2 + i \dfrac 4 5}^n$ ===
=== Example: $\paren {\cos \dfrac \pi {n + 1} + i \sin \dfrac \pi {n + 1} }^n$ ===
=== Example: $\tan i n$ ===
Category:Examples of Convergent Complex Sequences
Convergent Complex Sequence/Examples/((1 + i n) over (1 + n))^3

== Example of Convergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n = \paren {\dfrac {1 + i n} {1 + n} }^3$
Then:
:$\ds \lim_{n \mathop \to \infty} z_n = -i$
== Proof ==
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $1 \ \text {(i)}$
Category:Examples of Convergent Complex Sequences
Convergent Complex Sequence/Examples/(1 over 2 + i 4 over 5)^n

== Example of Convergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n = \paren {\dfrac 1 2 + i \dfrac 4 5}^n$
Then:
:$\ds \lim_{n \mathop \to \infty} z_n = 0$
== Proof ==
Thus $\cmod {z_n} \to 0$ and so $z_n \to 0$ as $n \to \infty$.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $1 \ \text {(iii)}$
Category:Examples of Convergent Complex Sequences
Convergent Complex Sequence/Examples/(3+in)^2 over n^2

== Example of Convergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n = \dfrac {\paren {3 + i n}^2} {n^2}$
Then:
:$\ds \lim_{n \mathop \to \infty} z_n = -1$
== Proof ==
== Sources ==
* : $\S 4.2$. Sequences: Example $\text {(i)}$
Category:Examples of Convergent Complex Sequences
Convergent Complex Sequence/Examples/(cos pi over n+1 + i sin pi over n+1)^n

== Example of Convergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n = \paren {\cos \dfrac \pi {n + 1} + i \sin \dfrac \pi {n + 1} }^n$
Then:
:$\ds \lim_{n \mathop \to \infty} z_n = -1$
== Proof ==
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $1 \ \text {(iv)}$
Category:Examples of Convergent Complex Sequences
Convergent Complex Sequence/Examples/(cos pi over n + i sin pi over n)^2n+1

== Example of Convergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n = \paren {\cos \dfrac \pi n + i \sin \dfrac \pi n}^{2 n + 1}$
Then:
:$\ds \lim_{n \mathop \to \infty} z_n = 1$
== Proof ==
== Sources ==
* : $\S 4.2$. Sequences: Example $\text {(v)}$
Category:Examples of Convergent Complex Sequences
Convergent Complex Sequence/Examples/tan i n

== Example of Convergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n = \tan i n$
Then:
:$\ds \lim_{n \mathop \to \infty} z_n = i$
== Proof ==
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $1 \ \text {(v)}$
Category:Examples of Convergent Complex Sequences
Convergent Complex Series/Examples

== Examples of Convergent Complex Series ==
=== Example: $\dfrac {\paren {-1}^n + i \cos n \theta} {n^2}$ ===
=== Example: $\dfrac 1 {n^2 - i n}$ ===
=== Example: $\dfrac {e^{i n} } {n^2}$ ===
=== Example: $\paren {\dfrac {2 + 3 i} {4 + i} }^n$ ===
Category:Examples of Convergent Series
Convergent Complex Series/Examples/((-1)^n + i cos n theta) over n^2

== Example of Convergent Complex Series ==
The series $\ds \sum_{n \mathop = 1}^\infty a_n$, where:
:$a_n = \dfrac {\paren {-1}^n + i \cos n \theta} {n^2}$
is convergent.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Convergent Series
Convergent Complex Series/Examples/((-1)^n + i cos n theta) over n^2/Proof 1

== Example of Convergent Complex Series ==
== Proof ==
Both of the terms on the are convergent real series.
Hence from Convergence of Series of Complex Numbers by Real and Imaginary Part, $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
== Sources ==
* : $\S 4.3$. Series: Example $\text{(i)}$
Category:Examples of Convergent Series
Convergent Complex Series/Examples/((-1)^n + i cos n theta) over n^2/Proof 2

== Example of Convergent Complex Series ==
== Proof ==
Thus $\ds \sum_{n \mathop = 1}^\infty \paren {\dfrac {\paren {-1}^n + i \cos n \theta} {n^2} }$ is absolutely convergent.
The result follows from Absolutely Convergent Series is Convergent: Complex Numbers.
== Sources ==
* : $\S 4.3$. Series: Example $\text{(i)}$
Category:Examples of Convergent Series
Convergent Complex Series/Examples/((2+3i) over (4+i))^n

== Example of Convergent Complex Series ==
The series $\ds \sum_{n \mathop = 1}^\infty a_n$, where:
:$a_n = \paren {\dfrac {2 + 3 i} {4 + i} }^n$
is convergent.
== Proof ==
Thus $\ds \sum_{n \mathop = 1}^\infty \paren {\dfrac {2 + 3 i} {4 + i} }^n$ is absolutely convergent by Sum of Infinite Geometric Sequence.
The result follows from Absolutely Convergent Series is Convergent: Complex Numbers.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $2 \ \text {(iv)}$
Category:Examples of Convergent Series
Convergent Complex Series/Examples/1 over n^2 - i n

== Example of Convergent Complex Series ==
The series $\ds \sum_{n \mathop = 1}^\infty a_n$, where:
:$a_n = \dfrac 1 {n^2 - i n}$
is convergent.
== Proof ==
Thus $\ds \sum_{n \mathop = 1}^\infty \dfrac 1 {n^2 - i n}$ is absolutely convergent.
The result follows from Absolutely Convergent Series is Convergent: Complex Numbers.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $2 \ \text {(i)}$
Category:Examples of Convergent Series
Convergent Complex Series/Examples/e^in over n^2

== Example of Convergent Complex Series ==
The series $\ds \sum_{n \mathop = 1}^\infty a_n$, where:
:$a_n = \dfrac {e^{i n} } {n^2}$
is convergent.
== Proof ==
Thus $\ds \sum_{n \mathop = 1}^\infty \dfrac {e^{i n} } {n^2}$ is absolutely convergent.
The result follows from Absolutely Convergent Series is Convergent: Complex Numbers.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $2 \ \text {(iii)}$
Category:Examples of Convergent Series
Convergent Generalized Sum of Positive Reals has Countably Many Non-Zero Terms

== Theorem ==
Let $\family {a_i}_{i \mathop \in I}$ be an $I$-indexed family of positive real numbers.
That is, let $a_i \in \R_{\ge 0}$ for all $i \in I$.
Suppose that $\ds \sum \set {a_i: i \in I}$ converges.
Then the set $I_{>0} := \set {i \in I: a_i > 0}$ is countable.
== Proof ==
For $\lambda \in \R$, let $I_{>\lambda} := \set {i \in I: a_i > \lambda}$.
Then as $\ds \sum \set {a_i: i \mathop \in I}$ converges, necessarily all of the sets $I_{> \frac 1 n}$ are finite.
However, we have that $\ds I_{>0} = \bigcup_{n \mathop = 1}^\infty I_{> \frac 1 n}$.
From Countable Union of Countable Sets is Countable, it follows that $I_{>0}$ is countable.
Category:Generalized Sums
Category:Real Analysis
Convergent Product Satisfies Cauchy Criterion

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ be convergent.
Then it satisfies Cauchy's criterion for products.
== Proof ==
Let $\epsilon > 0$.
Let $n_0 \in \N$ be such that $\ds \prod_{n \mathop = n_0}^\infty a_n$ converges to some $a \in \mathbb K \setminus \set 0$.
By Convergent Sequence is Cauchy Sequence, there exists $N_0 \ge n_0$ such that:
:$\ds \norm {\prod_{n \mathop = n_0}^k a_n - \prod_{n \mathop = n_0}^l a_n} \le \epsilon$
for $k, l \ge N_0$.
By Sequence Converges to Within Half Limit, there exists $N_1 \ge n_0$ such that:
:$\ds \norm {\prod_{n \mathop = n_0}^M a_n} \ge \frac {\norm a}2$
for $M \ge N_1$.
Let $N = \max \set {N_0, N_1}$.
For $N + 1 \le k \le l$:
Hence the result.
== Also see ==
* Uniformly Convergent Product Satisfies Uniform Cauchy Criterion
Category:Infinite Products
Category:Cauchy Sequences
Convergent Real Sequence/Examples

== Examples of Convergent Real Sequences ==
=== Example: $1 + \dfrac 1 n$ ===
=== Example: $\dfrac {n^2 - 1} {n^2 + 1}$ ===
=== Example: $\dfrac {2 n^3 - 3 n} {5 n^3 + 4 n^2 - 2}$ ===
=== Example: $\dfrac {n^3 + 5 n^2 + 2} {2 n^3 + 9}$ ===
=== Example: $\dfrac {x + x^n} {1 + x^n}$ ===
=== Example: $x_{n + 1} = {x_n}^2 + k$ ===
=== Example: $x_{n + 1} = \dfrac k {1 + x_n}$ ===
=== Example: $x_n = \sqrt {x_{n - 1} y_{n - 1} }$, $\dfrac 1 {y_n} = \dfrac 1 2 \paren {\dfrac 1 {x_n} + \dfrac 1 {y_{n - 1} } }$ ===
=== Example: $x_n = \dfrac {x_{n - 1} + x_{n - 2} } 2$ ===
=== Example: $\size {x_{n + 1} - x_n} \le \alpha^n$ ===
=== Example: $x_n = \sqrt {x_{n - 1} x_{n - 2} }$ ===
Category:Examples of Convergent Real Sequences
Convergent Real Sequence/Examples/1 plus Reciprocal of n

== Example of Convergent Real Sequence ==
The sequence $\sequence {a_n}_{n \mathop \ge 1}$ defined as:
:$a_n := 1 + \dfrac 1 n$
is convergent to the limit $1$ as $n \to \infty$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be given.
The requirement is to find a value of $N \in \R$ such that:
:$\forall n > N: \size {\paren {1 + \dfrac 1 n} - 1} < \epsilon$
But we have:
Hence the requirement is now to find a value of $N \in \R$ such that:
:$\forall n > N: \dfrac 1 n < \epsilon$
From Reciprocal Function is Strictly Decreasing:
:$\dfrac 1 n < \epsilon \implies n > \dfrac 1 \epsilon$
So choosing $N = \dfrac 1 \epsilon$ we have that:
:$\forall n > N: n > \dfrac 1 \epsilon$
and so:
:$\forall n > \dfrac 1 \epsilon: \size {\paren {1 + \dfrac 1 n} - 1} < \epsilon$
Hence the result.
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.5$: Example
Category:Examples of Convergent Real Sequences
Convergent Real Sequence/Examples/2 n^3 - 3 n over 5 n^3 + 4 n^2 - 2

== Example of Convergent Real Sequence ==
:$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {2 n^3 - 3 n} {5 n^3 + 4 n^2 - 2} } = \dfrac 2 5$
== Proof ==
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.9$: Example
Category:Examples of Convergent Real Sequences
Convergent Real Sequence/Examples/Arithmetic Mean of Previous 2 Terms

== Example of Convergent Real Sequence ==
Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:
:$x_n = \begin {cases} a & : n = 1 \\ b & : n = 2 \\ \dfrac {x_{n - 1} + x_{n - 2} } 2 & : n > 2 \end {cases}$
That is, beyond the first $2$ terms, each term is the arithmetic mean of the previous $2$ terms.
Then $\sequence {x_n}$ converges.
== Proof ==
Let $n > m$.
Then:
Let $\epsilon \in \R_{>0}$ be given.
Let $N$ be sufficiently large that:
:$\dfrac 1 {2^{N - 2} } \size {b - a} < \epsilon$
Then:
:$\forall n > N, m > N: \size {x_n - x_m} \le \dfrac 1 {2^{N - 2} } \size {b - a} < \epsilon$
Hence it is seen that $\sequence {x_n}$ is a Cauchy sequence.
Hence the result, by Cauchy's Convergence Criterion.
== Sources ==
* : $\S 5$: Subsequences: Cauchy sequences: Example $\S 5.20$
Category:Examples of Convergent Real Sequences
Convergent Real Sequence/Examples/Difference between Adjacent Terms Bounded by alpha^n

== Example of Convergent Real Sequence ==
Let $\alpha \in \R$ be a real number such that $0 < \alpha < 1$.
Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be a sequence in $\R$ with the property:
:$\size {x_{n + 1} - x_n} \le \alpha^n$
Then $\sequence {x_n}$ is a Cauchy sequence and hence converges.
== Proof ==
Let $n > m$.
Then:
Let $\epsilon \in \R_{>0}$ be given.
Let $N$ be sufficiently large that:
:$\dfrac {\alpha^m} {1 - \alpha} < \epsilon$
This is always possible, as shown by Sequence of Powers of Number less than One.
Hence the result, by Cauchy's Convergence Criterion.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.21 \ (1)$
Category:Examples of Convergent Real Sequences
Convergent Real Sequence/Examples/Term is Geometric Mean of Preceding Two Terms

== Example of Convergent Real Sequence ==
Let $a, b \in \R_{>0}$ be (strictly) positive real numbers such that $a \le b$.
Let $a, b \in \R_{>0}$ be such that $a \le p \le q \le b$.
Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:
:$x_n = \begin {cases} p & : n = 1 \\ q & : n = 2 \\ \sqrt{x_{n - 1} x_{n - 2} } & : n > 2 \end {cases}$
That is, beyond the first $2$ terms, each term is the geometric mean of the previous $2$ terms.
Then $\sequence {x_n}$ converges.
== Proof ==
First note that:
:$a \le x_i \le b$ for $i \in \set {1, 2}$.
It can be shown by Proof by Mathematical Induction that:
:$\forall n \in \N_{>0}: a \le x_n \le b$
Then:
and:
That is:
:$\dfrac a b \le \dfrac {x_{n + 1} } {x_n} \le \dfrac b a$
Then we have:
Let $n > m$ be arbitrary. Then:
Therefore $\sequence {x_n}$ is a Cauchy sequence and hence converges.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.21 \ (2)$
 Category:Examples of Convergent Real Sequences
Convergent Real Sequence/Examples/n^2 - 1 over n^2 + 1

== Example of Convergent Real Sequence ==
:$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n^2 - 1} {n^2 + 1} } = 1$
== Proof ==
Let $\epsilon \in \R_{>0}$ be given.
The requirement is to find a value of $N \in \R$ such that:
:$\forall n > N: \size {\dfrac {n^2 - 1} {n^2 + 1} - 1} < \epsilon$
But we have:
Hence the requirement is now to find a value of $N \in \R$ such that:
:$\forall n > N: \dfrac 2 {n^2 + 1} < \epsilon$
From Reciprocal Function is Strictly Decreasing:
:$\dfrac 2 {n^2 + 1} < \epsilon \implies n^2 + 1 > \dfrac 2 \epsilon$
So choosing $N = \paren {\dfrac 2 \epsilon}^{1/2}$ we have that:
:$\forall n > N: n > \dfrac 1 \epsilon$
We have demonstrated that for $\epsilon \in \R_{>0}$ there exists $N \in \R$, that is: $N = \paren {\dfrac 2 \epsilon}^{1/2}$, such that:
:$\forall n > N: \size {\dfrac {n^2 - 1} {n^2 + 1} - 1} < \epsilon$
Hence the result:
:$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n^2 - 1} {n^2 + 1} } = 1$
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.6 \ (1)$
Category:Examples of Convergent Real Sequences