nyuuzyou/wikis huggingface dataset:
Convergence in Direct Product Norm

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces.
Let $X \times Y$ be a direct product of vector spaces $X$ and $Y$ together with induced component-wise operations.
Let $\norm {\, \cdot \,}_{X \times Y}$ be the direct product norm.
Let $\sequence {\tuple {x_n, y_n} }_{n \mathop \in \N}$ be a sequence in $X \times Y$.
Then: 
:$\tuple {x_n, y_n}$ converges to $\tuple {x, y}$ in $X \times Y$
:
:$x_n$ converges to $x$ in $X$
and:
:$y_n$ converges to $y$ in $Y$.
 
== Proof ==
=== Neccessary Condition ===
Suppose that:
:$\tuple {x_n, y_n}$ converges to $\tuple {x, y}$ in $X \times Y$
Then, for each $\epsilon > 0$, there exists $N \in \N$ such that: 
:$\ds \norm {\tuple {x_n, y_n} - \tuple {x, y} }_{X \times Y} < \epsilon$
for all $n \ge N$.
This is equivalent to: 
:$\ds \norm {\tuple {x_n - x, y_n - y} }_{X \times Y} < \epsilon$
by the definition of induced component-wise operations.
From the definition of the direct product norm, this is equivalent to: 
:$\ds \map \max {\norm {x_n - x}_X, \norm {y_n - y}_Y} < \epsilon$
This is equivalent to: 
:$\norm {x_n - x}_X < \epsilon$
and:
:$\norm {y_n - y}_Y < \epsilon$
for $n \ge N$. 
That is: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges to $x$ and $\sequence {y_n}_{n \mathop \in \N}$ converges to $y$.
=== Sufficient Condition ===
Suppose that: 
:$x_n$ converges to $x$ in $X$
and:
:$y_n$ converges to $y$ in $Y$.
Let $\epsilon > 0$.
Then we can pick $N_1 \in \N$ such that: 
:$\norm {x_n - x}_X < \epsilon$
for $n \ge N_1$.
Similarly, we can pick $N_2 \in \N$ such that: 
:$\norm {y_n - y}_Y < \epsilon$
Let:
:$N = \max \set {N_1, N_2}$
Then, for $n \ge N$, we have: 
:$\norm {x_n - x}_X < \epsilon$
and:
:$\norm {y_n - y}_Y < \epsilon$
so that: 
:$\map \max {\norm {x_n - x}_X, \norm {y_n - y}_Y} < \epsilon$
From the definition of the direct product norm, we have: 
:$\norm {\tuple {x_n - x, y_n - y} }_{X \times Y} < \epsilon$
From the definition of induced component-wise operations, we have: 
:$\norm {\tuple {x_n, y_n} - \tuple {x, y} }_{X \times Y} < \epsilon$
for all $n \ge N$.
So:
:$\tuple {x_n, y_n}$ converges to $\tuple {x, y}$ in $X \times Y$
Category:Direct Product of Vector Spaces
Category:Convergent Sequences (Normed Vector Spaces)
Convergence in Indiscrete Space

== Theorem ==
Let $\struct {S, \set {S, \O} }$ be an indiscrete space.
Let $\sequence {x_n}$ be any sequence in $S$.
Then $\sequence {x_n}$ converges to any point $x$ of $S$.
== Proof ==
For any open set $U \subseteq S$ such that $x \in U$, we must have $U = S$.
Hence:
:$\forall n \ge 1: x_n \in U$
The result follows from the definition of a convergent sequence in a topological space.
== Sources ==
* : $4$: The Hausdorff condition: $4.1$: Motivation: Example $4.1.3$
Category:Indiscrete Topology
Category:Convergence
Convergence in Mean implies Convergence in Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $f : X \to \R$ be a $\Sigma$-measurable function.
For each $n \mathop \in \N$, let $f_n : X \to \R$ be a $\Sigma$-measurable function.
Then:
:if $\sequence {f_n}_{n \mathop \in \N}$ converges in mean to $f$, it converges in measure to $f$. 
== Proof ==
From Pointwise Difference of Measurable Functions is Measurable:
:$f_n - f$ is $\Sigma$-measurable for each $n \in \N$. 
Let $\epsilon > 0$ be a real number.
From Markov's Inequality, we then have: 
:$\ds \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon} } \le \frac 1 \epsilon \int \size {f_n - f} \rd \mu$
Since $\sequence {f_n}_{n \mathop \in \N}$ converges in mean to $f$, we have: 
:$\ds \lim_{n \mathop \to \infty} \frac 1 \epsilon \int \size {f_n - f} \rd \mu = 0$
So, from the Squeeze Theorem, we have: 
:$\ds \lim_{n \mathop \to \infty} \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon} } = 0$
for each $\epsilon > 0$.
So:
:$\sequence {f_n}_{n \mathop \in \N}$ converges in measure to $f$.
Category:Convergence in Mean
Category:Convergence in Measure
Convergence in Measure Implies Convergence a.e. of Subsequence

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $D \in \Sigma$. 
Let $f : D \to \R$ be a $\Sigma$-measurable function.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-measurable functions $f_n : D \to \R$ such that: 
:$\sequence {f_n}_{n \mathop \in \N}$ converges in measure to $f$. 
Then there is a subsequence $\sequence {f_{n_k} }_{k \mathop \in \N}$ of $\sequence {f_n}_{n \mathop \in \N}$ that converges a.e. to $f$.
 
== Proof ==
For each $n, k \ge 1$, define:
:$\ds B_{n, k} = \set {x \in X : \size {\map {f_n} x - \map f x} > \frac 1 k}$
Since $\sequence {f_n}_{n \mathop \in \N}$ converges in measure to $f$, we have: 
:$\ds \lim_{n \mathop \to \infty} \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \frac 1 k} } = 0$
Then from the definition of convergence, for each $k$ we can select $n_k$ such that: 
:$\ds \map \mu {\set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \frac 1 k} } < \frac 1 {2^k}$ 
That is: 
:$\ds \map \mu {B_{n_k, k} } < \frac 1 {2^k}$ 
We then have: 
By the Borel-Cantelli Lemma, we therefore have: 
:$\ds \map \mu {\limsup_{k \mathop \to \infty} B_{n_k, k} } = 0$
By Equivalence of Definitions of Limit Superior of Sequence of Sets:
:$\map \mu {\set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k \text { for infinitely many } k} } = 0$
Let: 
:$N = \set {x \in X : \size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k \text { for infinitely many } k}$ 
Then, for $x \not \in N$, we have: 
:$\size {\map {f_{n_k} } x - \map f x} > \dfrac 1 k$ holds for only finitely many $k$.
That is, there exists $K \in \N$ such that: 
:$\size {\map {f_{n_k} } x - \map f x} \le \dfrac 1 k$ for $k \ge K$.
Then from the Squeeze Theorem, we have: 
:$\size {\map {f_{n_k} } x - \map f x} \to 0$
so:
:$\map {f_{n_k} } x \to \map f x$ 
So if $x \notin N$: 
:$\map {f_{n_k} } x \to \map f x$
Since $\map \mu N = 0$: 
:$\sequence {f_{n_k}}_{n_k \mathop \in \N}$ converges a.e. to $f$.
== Sources ==
* : $1.1.2$: Convergence in Measure
Category:Convergence Almost Everywhere
Convergence in Norm Implies Convergence in Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \R, p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a sequence of $p$-integrable functions.
Also, let $f: X \to \R$ be a $p$-integrable function.
Suppose that $f_n$ converges in norm to $f$ (in the $L^p$-norm).
Then $f_n$ converges in measure to $f$ (in $\mu$).
That is:
:$\ds \operatorname {\LL^{\textit p}-\!\lim\,} \limits_{n \mathop \to \infty} f_n = f \implies \operatorname {\mu-\!\lim\,} \limits_{n \mathop \to \infty} f_n = f$
== Proof ==
Let $\epsilon > 0$.
Then:
== Sources ==
* : $16.4 \ \text{(i)}$
Category:Measure Theory
Convergence in Normed Dual Space implies Weak-* Convergence

== Theorem ==
 
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a convergent sequence in $X^\ast$.
Then $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* $27.4$: Weak-$\ast$ Convergence
Category:Weak-* Convergence (Normed Vector Spaces)
Category:Convergence in Normed Dual Space implies Weak-* Convergence
Convergence in Normed Dual Space implies Weak-* Convergence/Proof 1

== Theorem ==
== Proof ==
From Convergent Sequence in Normed Vector Space is Weakly Convergent, $\sequence {f_n}_{n \mathop \in \N}$ converges weakly.
From Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent, $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$.
Category:Convergence in Normed Dual Space implies Weak-* Convergence
Convergence in Normed Dual Space implies Weak-* Convergence/Proof 2

== Theorem ==
== Proof ==
Let $f \in X^\ast$ be the limit of $\sequence {f_n}_{n \mathop \in \N}$, i.e.:
:$\norm {f_n - f}_{X^\ast} \stackrel{n \to \infty}{\longrightarrow} 0$
Thus, for each $x \in X$:
Category:Convergence in Normed Dual Space implies Weak-* Convergence
Convergence in Sigma-Finite Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a $\sigma$-finite measure space.
Let $\sequence {f_n}_{n \mathop \in \N}: f_n: X \to \R$ be a sequence of measurable functions.
Also, let $f, g: X \to \R$ be measurable functions.
Suppose that $f_n$ converges in measure to both $f$ and $g$ (in $\mu$).
Then $f$ and $g$ are equal $\mu$-almost everywhere.
== Proof ==
As $\mu$ is $\sigma$-finite, there is an exhausting sequence $\sequence {E_m}_m$ in $\Sigma$ such that:
:$\forall m : \map \mu {E_m} < \infty$
By triangle inequality:
:$\forall x \in X :\quad \cmod {\map f x - \map g x} \le \cmod {\map f x - \map {f_n} x} + \cmod {\map {f_n} x - \map g x}$
For all $m, k \in \N_{>0}$, since:
:$\ds \forall n \in \N :\quad \set {\cmod {f - g} > \frac 1 k } \subseteq \set {\cmod {f - f_n} > \frac 1 {2 k} } \cup \set {\cmod {f_n - g} > \frac 1 {2 k} }$
we have:
Thus:
:$\ds \forall m, k \in \N_{>0} : \map \mu {\set {\cmod {f - g} > \frac 1 k} \cap E_m} = 0$
By Monotone Convergence Theorem:
:$\ds \forall k \in \N_{>0} : \map \mu {\set {\cmod {f - g} > \frac 1 k} } = \lim_{m \mathop \to \infty} \map \mu {\set {\cmod {f - g} > \frac 1 k} \cap E_m} = 0$
Now, observe:
:$\ds \set {\cmod {f - g} > 0} = \bigcup_{k \mathop = 1}^\infty \set {\cmod {f - g} > \frac 1 k}$
Therefore:
== Sources ==
* : $16.5$
Category:Measure Theory
Convergence of Complex Conjugate of Convergent Complex Sequence

== Theorem ==
Let $z \in \C$. 
Let $\sequence {z_n}_{n \mathop \in \N}$ be a complex sequence converging to $z$. 
Then: 
:$\overline {z_n} \to \overline z$
== Proof ==
Let $\epsilon > 0$.
Since $z_n \to z$, from the definition of convergence, we can find $N \in \N$ such that: 
:$\cmod {z_n - z} < \epsilon$ 
From Complex Modulus equals Complex Modulus of Conjugate, we have: 
:$\cmod {\overline {z_n - z} } = \cmod {z_n - z}$
From Difference of Complex Conjugates, we have: 
:$\cmod {z_n - z} = \cmod {\overline {z_n} - \overline z}$
So we have: 
:$\cmod {\overline {z_n} - \overline z} < \epsilon$ 
for each $n \ge N$.
Since $\epsilon$ was arbitrary, we have: 
:$\overline {z_n} \to \overline z$
Category:Complex Conjugates
Category:Convergent Complex Sequences
Convergence of Complex Sequence in Polar Form

== Theorem ==
Let $z \ne 0$ be a complex number with modulus $r$ and argument $\theta$.
Let $\sequence {z_n}$ be a sequence of nonzero complex numbers.
Let $r_n$ be the modulus of $z_n$ and $\theta_n$ be an argument of $z_n$.
Then $z_n$ converges to $z$ the following hold:
:$(1): \quad r_n$ converges to $r$
:$(2): \quad$ There exists a sequence $\sequence {k_n}$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$.
=== Corollary ===
== Proof ==
Suppose $r_n \to r$ and $\theta_n + 2 k_n \pi \to \theta$.
We have, by Complex Modulus of Difference of Complex Numbers:
Because Cosine Function is Continuous:
:$\map \cos {\theta_n + 2 k_n \pi - \theta} \to 1$
It follows that:
:$\cmod {z_n - z}^2 \to 0$
Conversely, suppose $z_n \to z$.
By Modulus of Limit, $r_n \to r$.
We have, by Complex Modulus of Difference of Complex Numbers:
By Convergence of Cosine of Sequence, there exists a sequence $\sequence {k_n}$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$.
Category:Complex Analysis
Convergence of Complex Sequence in Polar Form/Corollary

== Corollary to Convergence of Complex Sequence in Polar Form ==
Let $\left\langle{z_n}\right\rangle$ be a sequence of nonzero complex numbers.
Let $z \ne 0$ be a complex number with modulus $r$ and argument $\theta$.
Let $I$ be a real interval of length at most $2 \pi$ that contains $\theta$.
Suppose $\theta$ is not an endpoint of $I$.
Suppose each $z_n$ admits an argument $\theta_n \in I$.
Let $r_n$ be the modulus of $z_n$.
Then $z_n$ converges to $z$ $r_n$ converges to $r$ and $\theta_n$ converges to $\theta$.
== Proof ==
Suppose $r_n \to r$ and $\theta_n \to \theta$.
Then by Convergence of Complex Sequence in Polar Form, $z_n \to z$.
Conversely, suppose $z_n \to z$.
By Convergence of Complex Sequence in Polar Form, we have:
:$(1): \quad r_n \to r$
:$(2): \quad$ There exists a sequence $\left\langle{k_n}\right\rangle$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$.
It remains to be proved that $\theta_n \to \theta$.
Let $N \in \N$ such that $\left\vert{\theta_N + 2 k_N \pi - \theta}\right\vert \le \pi / 2$ for all $n \ge N$.
By the Triangle Inequality for Real Numbers:
:$\left\vert{2 k_n \pi - 2 k_N \pi}\right\vert \le \left\vert{\theta_n + 2 k_n \pi - \theta}\right\vert + \left\vert{\theta_N + 2 k_N \pi - \theta}\right\vert \le \pi$
for all $n \ge N$.
Thus $\left\vert{k_n - k_N}\right\vert \le 1 / 2$, so $k_n = k_N$ for all $n \ge N$.
By the Triangle Inequality for Real Numbers:
:$\left\vert{2 \pi k_N}\right\vert \le \left\vert{\theta_n - \theta}\right\vert + \left\vert{\theta_n + 2 \pi k_N - \theta}\right\vert$
for all $n \in \N$.
Because $\theta_n \in I$ and $\theta$ is not an endpoint of $I$:
:$\left\vert{\theta_n - \theta}\right\vert < 2 \pi$
for all $n \in \N$.
Because $\theta_n + 2 \pi k_N - \theta \to 0$:
:$\left\vert{2 \pi k_N}\right\vert < 2 \pi$
Thus $k_n = 0$ for all $n \ge N$.
Thus $\theta_n \to \theta$.
Category:Complex Analysis
Convergence of Dirichlet Series with Bounded Coefficients

== Theorem ==
Let $\sequence {a_n}_{n \mathop \in \N}$ be a bounded sequence in $\C$.
Then the Dirichlet series:
:$\ds \map f s = \sum_{n \mathop \ge 1} a_n n^{-s}$
converges absolutely and locally uniformly to an analytic function on $\map \Re s > 1$.
== Proof ==
By Exponential is Entire, the partial sums:
:$\ds \map {f_N} s = \sum_{n \mathop = 1}^N a_n n^{-s}$
are analytic.
So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show locally uniform convergence.
Let $B$ be a bound for the $a_n$:
:$\forall n \in \N: \size {a_n} \le B$
Let $D$ be any open subset of $\map \Re s > 1$.
So for some $\kappa > 0$:
:$\forall s \in D: \map \Re s \ge 1 + \kappa$
Now:
which we know to be finite (!).
Category:Analytic Number Theory
Convergence of Dirichlet Series with Bounded Partial Sums

== Theorem ==
Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence in $\C$.
Suppose that there exists $B > 0$ such that for all $n, m \in \N$:
:$\ds \size {\sum_{k \mathop = m}^n a_n} \le B$
Then the Dirichlet series:
:$\ds \map f s = \sum_{n \mathop \ge 1} a_n n^{-s}$
converges locally uniformly to an analytic function on $\map \Re s > 0$.
== Proof ==
By Exponential is Entire, the partial sums:
:$\ds \map {f_N} s = \sum_{n \mathop = 1}^N a_n n^{-s}$
are analytic.
So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show locally uniform convergence.
For $0 < A < \pi / 2$, $\delta > 0$ we let:
:$D_{A, \delta} = \set {s \in \C : -A < \map \arg s < A, \map \Re s > \delta}$
Then for any $s \in \C$ such that $\map \Re s > 0$, we can choose $A$, $\delta$ such that $s \in D_{A, \delta}$.
So it is sufficient to prove locally uniform convergence in this region.
Also note that if $\lambda_n = \log n$, then:
:$\ds \map f s = \sum_{n \mathop \ge 1} a_n e^{-\lambda_n s}$
By Abel's Lemma: Formulation 2, for $N, M \in \N$ we have:
:$\ds \sum_{n \mathop = M}^N a_n e^{-\lambda_n s} = \sum_{n \mathop = M}^{N - 1} \paren {\sum_{k \mathop = M}^n a_n} \paren {e^{-\lambda_n s} - e^{-\lambda_{n + 1} s} } + e^{-\lambda_N s} \paren {\sum_{n \mathop = M}^N a_n}$
Let $\epsilon > 0$ be arbitrary, and choose $N_0 \in \N$ such that:
:$\ds \forall n \ge N_0: \size {e^{-\lambda_n s} } < \epsilon$
Now for $N, M \ge N_0$, by the above we obtain:
:$(1): \quad \ds \size {\sum_{n \mathop = M}^N a_n e^{-\lambda_n s} } \le B \sum_{n \mathop = M}^{N - 1} \size {e^{-\lambda_n s} - e^{-\lambda_{n + 1} s} } + B \epsilon$
For any $\alpha, \beta \in \R$ we have:
Therefore:
Finally letting $\map \Im s = t$:
:$\dfrac {\size s} \sigma = \dfrac {\sqrt {\sigma^2 + t^2} } \sigma = \sqrt {1 + \dfrac {t^2} {\sigma^2} }$
and:
:$\dfrac {\pi} 2 > A \ge \map \arg s = \map \arctan {\dfrac t \sigma}$
so $\dfrac t \sigma < \tan A$ is bounded uniformly in $D_{A, \delta}$.
Thus letting $N \to \infty$, we have shown that:
:$\ds \size {\sum_{n \mathop = M}^\infty a_n e^{-\lambda_n s} } \to 0$
as $M \to \infty$ uniformly in $D_{A, \delta}$.
Category:Analytic Number Theory
Convergence of Generalized Sum of Complex Numbers

== Theorem ==
Let $\family {z_j}_{j \mathop \in I}$ be an $I$-indexed family of complex numbers.
That is, let $z_j \in \C$ for all $j \in I$.
Let $\map \Re {z_j}$ and $\map \Im {z_j}$ denote the families of real and imaginary parts of the family $z_j$.
Then the following are equivalent:
:$(1): \quad \ds \sum_{j \mathop \in I} z_j$ converges to $z \in \C$
:$(2): \quad \ds \sum_{j \mathop \in I} \map \Re {z_j}, \sum_{j \mathop \in I} \map \Im {z_j}$ converge to $\map \Re z, \map \Im z \in \R$, respectively
=== Corollary ===
== Proof ==
=== $(2)$ implies $(1)$ ===
By Generalized Sum is Linear, the stated convergences lead to:
=== $(1)$ implies $(2)$ ===
Statement $(1)$, according to the definition of convergence, amounts to the following:
For every $\epsilon > 0$, there exists a finite $G \subseteq I$ such that:
:For every finite $F \subseteq I$ with $G \subseteq F$:
::$\ds \cmod {z - \sum_{j \mathop \in F} z_j} < \epsilon$
Now suppose that for $\epsilon > 0$, $G$ and $F$ are as above. Then observe that:
Hence, by Square of Real Number is Non-Negative, both of the terms on the are smaller than $\epsilon^2$.
It follows that, taking square roots, $G$ satisfies, for any finite $F \supseteq G$:
:$\ds \size {\map \Re z - \sum_{j \mathop \in F} \map \Re {z_j} } < \epsilon$
:$\ds \size {\map \Im z - \sum_{j \mathop \in F} \map \Im {z_j} } < \epsilon$
As $\epsilon > 0$ was arbitrary, using the definition of convergence, this implies precisely that:
:$\ds \sum_{j \mathop \in I} \map \Re {z_j}, \sum_{j \mathop \in I} \map \Im {z_j}$ converge to $\map \Re z, \map \Im z \in \R$, respectively.
Hence, $(1)$ is shown to imply $(2)$.
== Also see ==
* Generalized Sum is Linear, of which this is a partial converse.
Category:Generalized Sums
Category:Complex Numbers
Convergence of Generalized Sum of Complex Numbers/Corollary

== Corollary to Convergence of Generalized Sum of Complex Numbers ==
Let $\family {z_j}_{j \mathop \in I}$ be an $I$-indexed family of complex numbers.
That is, let $z_j \in \C$ for all $j \in I$.
Suppose that $\ds \sum_{j \mathop \in I} z_j$ converges to $z \in \C$.
Then $\ds \sum_{j \mathop \in I} \overline {z_j}$ converges to $\overline z$, where $\overline z$ denotes the complex conjugate of $z$.
Here, the $\sum$ denote generalized sums.
== Proof ==
Using Convergence of Generalized Sum of Complex Numbers, one has:
:$\ds \sum_{j \mathop \in I} \map \Re {z_j} = \map \Re z$
:$\ds \sum_{j \mathop \in I} \map \Im {z_j} = \map \Im z$
Now, observe that, from the definition of complex conjugate:
Category:Generalized Sums
Category:Complex Numbers
Convergence of Limsup and Liminf

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let the limit superior of $\sequence {x_n}$ be $\overline l$.
Let the limit inferior of $\sequence {x_n}$ be $\underline l$.
Then $\sequence {x_n}$ converges to a limit $l$ $\overline l = \underline l = l$.
Hence a bounded real sequence converges all its convergent subsequences have the same limit.
== Proof ==
=== Sufficient Condition ===
First, suppose that $\overline l = \underline l = l$.
Let $\epsilon > 0$.
By Terms of Bounded Sequence Within Bounds:
:$\exists N_1: \forall n > N_1: x_n < l + \epsilon$
Similarly:
:$\exists N_2: \forall n > N_2: x_n > l - \epsilon$
So take $N = \max \set {N_1, N_2}$.
If $n > N$, both the above inequalities hold at the same time.
So $l - \epsilon < x_n < l + \epsilon$ and so by Negative of Absolute Value:
:$\size {x_n - l} < \epsilon$
Thus $x_n \to l$ as $n \to \infty$.
=== Necessary Condition ===
Now suppose that $\sequence {x_n}$ converges to a limit $l$.
Then by Limit of Subsequence equals Limit of Real Sequence, all subsequences have a limit of $l$ and the result follows.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $5$
* : $\S 5$: Subsequences: Exercise $\S 5.15 \ (5)$
* : $\S 8$
Category:Limits of Sequences
Category:Convergence
Convergence of Modulus of Convergent Complex Sequence

== Theorem ==
Let $\sequence {z_n}$ be a sequence in $\C$.
Let $\sequence {z_n}$ converge to a value $c \in \C$.
Let $\cmod z$ denote the modulus of a complex number $z$.
Then:
:$\sequence {\cmod {z_n} }$ converges to a value $\cmod c$.
== Proof ==
By definition of convergent complex sequence:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \cmod {z_n - c} < \epsilon$
From the Reverse Triangle Inequality:
:$\size {\cmod x - \cmod y} \le \cmod {x - y}$
and the result follows.
== Sources ==
* : $\S 4.2$. Sequences: Corollary.
Category:Convergent Complex Sequences
Convergence of Odd and Even Subsequences to Same Limit

== Theorem ==
Let $\sequence {s_n}$ be a real sequence.
Let the subsequences $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to the same limit $l$.
Then $\sequence {s_n}$ also converges to the same limit $l$.
== Proof ==
Suppose that $\sequence {s_n}$ converge to a limit.
Then from Limit of Subsequence equals Limit of Real Sequence, $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to the same limit.
They do so converge, and that limit is $l$.
So, if $\sequence {s_n}$ converges, it converges to the limit $l$.
 $\sequence {s_n}$ does not actually converge to a limit.
Then it is not the case that:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: k > N \implies \size {s_k - l} < \epsilon$
That is:
:$\exists \epsilon \in \R_{>0}: \forall N \in \R_{>0}: \exists k > N: \size {s_k - l} > \epsilon$
Either:
:$k = 2 n$ for some $n \in \N$
or:
:$k = 2 n + 1$ for some $n \in \N$
In the first case that means $\sequence {s_{2 n} }$ does not converge to $l$.
In the second case that means $\sequence {s_{2 n + 1} }$ does not converge to $l$.
Both cases contradict the supposition that $\sequence {s_{2 n} }$ and $\sequence {s_{2 n + 1} }$ both converge to $l$.
Hence by Proof by Contradiction it must be the case that $\sequence {s_n}$ converges to a limit.
That limit has been shown to be $l$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 12$
Category:Subsequences
Convergence of P-Series

== Theorem ==
Let $p \in \C$ be a complex number.
=== Absolute Convergence if $\map \Re p > 1$ ===
=== Divergence if $0 <\map \Re p \le 1$ ===
=== Real Case ===
== Also see ==
The mapping $\ds p \mapsto \sum_{n \mathop = 1}^\infty n^{-p}$ is well-known as the Riemann zeta function.
== Sources ==
* 
* 
Category:Convergence of P-Series
Category:Convergence Tests
Category:P-Series
Convergence of P-Series/Absolute Convergence if Real Part of p Greater than 1

== Theorem ==
Let $p$ be a complex number.
Let $\map \Re p > 1$.
Then the $p$-series:
:$\ds \sum_{n \mathop = 1}^\infty n^{-p}$
converges absolutely.
== Proof ==
=== Lemma ===
Since $x > 1$ it follows that $1 - x < 0$.
Thus $P^{1 - x} \to 0$ as $P \to \infty$.
Setting $x - 1 = \delta >0$, this limit is:
:$\ds -\frac 1 {\delta} \lim_{t \mathop \to \infty} \frac 1 {t^\delta} = 0$ 
Hence the integral is just $\dfrac 1 {1 - x}$ (that is, convergent) and so the sum converges as well.
Since the terms of the sum were positive everywhere, it is absolutely convergent and hence so is:
:$\ds \sum_{n \mathop = 1}^\infty n^{-p}$
== 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/Divergence if p between 0 and 1

== Theorem ==
Let $p$ be a complex number.
Let $0 < \map \Re p \le 1$.
Then the $p$-series:
:$\ds \sum_{n \mathop = 1}^\infty n^{-p}$
diverges.
== Proof ==
=== Lemma ===
Hence, the convergence of the $p$-series is dependent on the convergence of: 
:$\ds \lim_{t \mathop \to \infty} \frac {t^{1 - x} } {1 - x}$
Suppose $0 < x < 1$.
Then:
The special case of $x = 1$ is covered by Integral of Reciprocal is Divergent.
Hence the result from the Cauchy Integral Test.
Category:Convergence of P-Series