hcju/proofwikiqa huggingface dataset:
There are $2$ cases to consider. First, suppose that every set $\set {x_n: n > N}$ has a maximum. If this is the case, we can find a sequence $n_r \in \N$ such that: :$\displaystyle x_{n_1} = \max \set {x_n: n > 1}$ :$\displaystyle x_{n_2} = \max \set {x_n: n > n_1}$ :$\displaystyle x_{n_3} = \max \set {x_n: n > n_2}$ and so on. From the method of construction, $n_1 < n_2 < n_3 < \cdots$, so at each stage we are taking the maximum of a subset of the previous set. At each stage, the previous maximum has already been taken as the previous term in the sequence. Thus, $\sequence {x_{n_r} }$ is a decreasing subsequence of $\sequence {x_n}$. {{qed|lemma}} Second, suppose it is ''not'' true that ''every'' set $\set {x_n: n > N}$ has a maximum. Then there will exist some $N_1$ such that $\set {x_n: n > N_1}$ has no maximum. So it follows that, given some $x_m$ with $m > N_1$, we can find an $x_n$ following that $x_m$ such that $x_n > x_m$. (Otherwise, the biggest of $x_{N_1 + 1}, \ldots, x_m$ would be a maximum for $\set {x_n: n > N_1}$.) So, we define $x_{n_1} = x_{N_1 + 1}$. Then $x_{n_2}$ can be the first term after $x_{n_1}$ such that $x_{n_2} > x_{n_1}$. Then $x_{n_3}$ can be the first term after $x_{n_2}$ such that $x_{n_3} > x_{n_2}$. And so on. Thus we get an increasing subsequence of $\sequence {x_n}$. {{qed|lemma}} There exist only these two possibilities. From each one we get a subsequence that is monotone: :one is decreasing :one is increasing. We can of course choose instead to investigate whether every set $\set {x_n: n > N}$ has a minimum. The same conclusion will be reached. {{qed}}
A coastal town in Spain has an infinite row of hotels along a road leading down to the beach. The tourist bureau which rates these hotels has a special designation for any hotel having a view of the sea. Any hotel which is at least as tall as the rest of the hotels on the road to the sea receives this view-rating. Starting with the hotel farthest from the sea, let $x_1, x_2, \ldots$ denote the heights of the hotels as we travel along the road to the beach. One possibility is that an infinite number of hotels are view-rated. In this case the heights of these hotels form a decreasing (i.e. nonincreasing) subsequence of the original sequence of heights. The other possibility is that only a finite number of hotels are view-rated. In this case we can obtain an increasing subsequence as follows. Walk along the road to the beach past all the view-rated hotels. Let $n_1$ be the index of the next hotel. Since it does not have a view of the sea, there is a taller hotel nearer the shore which blocks the view. Let $n_2$ be the index of that taller hotel. Since it is not view-rated, it does not have a view of the sea. Thus there must be an even taller hotel nearer the shore which blocks the view. In this way we generate a list of indices $n_1 < n_2 < n_3 \cdots$ of taller and taller hotels. This gives an increasing subsequence $x_{n_1}, x_{n_2}, x_{n_3}, \ldots$ of the original sequence. {{qed}}
Let $\sequence {x_n}$ be a bounded sequence in $\R$. By the Peak Point Lemma, $\sequence {x_n}$ has a monotone subsequence $\sequence {x_{n_r} }$. Since $\sequence {x_n}$ is bounded, so is $\sequence {x_{n_r} }$. Hence, by the Monotone Convergence Theorem (Real Analysis), the result follows. {{qed}}
Let $\left \langle {x_n} \right \rangle_{n \in \N}$ be a bounded sequence in $\R$. By definition there are real numbers $c, C \in \R$ such that $c < x_n < C$. Then at least one of the sets: :$\left\{{x_n : c < x_n < \dfrac{c + C} 2 }\right\}, \left\{{x_n : \dfrac{c + C} 2 < x_n < C }\right\}, \left\{{x_n : x_n = \dfrac{c + C} 2 }\right\}$ contains infinitely many elements. If the set $\left\{{x_n : x_n = \dfrac{c + C} 2 }\right\}$ is infinite there's nothing to prove. If this is not the case, choose the first element from the infinite set, say $x_{k_1}$. Repeat this process for $\left \langle {x_n} \right \rangle_{n > k_1}$. As a result we obtain subsequence $\left \langle {x_{k_n}} \right \rangle_{n \in \N}$. By construction $\left \langle {x_{k_n}} \right \rangle_{n \in \N}$ is a Cauchy sequence and therefore converges. {{qed}}
From the Bolzano-Weierstrass Theorem: :$L \ne \O$ From Lower and Upper Bounds for Sequences, $L$ is a bounded subset of $\R$. Thus $L$ does have a supremum and infimum in $\R$. The object of this proof is to confirm that: :$\overline l := \map \sup L \in L$ and: :$\underline l := \map \inf L \in L$ that is, that these points do actually belong to $L$. First we show that $\overline l \in L$. To do this, we show that: :$\exists \sequence {x_{n_r} }: x_{n_r} \to \overline l$ as $n \to \infty$ where $\sequence {x_{n_r} }$ is a subsequence of $\sequence {x_n}$. Let $\epsilon > 0$. Then $\dfrac \epsilon 2 > 0$. Since $\overline l = \map \sup L$, and therefore by definition the ''smallest'' upper bound of $L$, $\overline l - \dfrac \epsilon 2$ is ''not'' an upper bound of $L$. Hence: :$\exists l \in L: \overline l \ge l > \overline l - \dfrac \epsilon 2$ Therefore: :$\sequence {l - \overline l} < \dfrac \epsilon 2$ Now because $l \in L$, we can find $\sequence {x_{m_r} }$, a subsequence of $\sequence {x_n}$, such that $x_{m_r} \to l$ as $n \to \infty$. So: :$\exists R: \forall r > R: \size {x_{m_r} - \overline l} < \dfrac \epsilon 2$ So, for any $r > R$: {{begin-eqn}} {{eqn | l = \size {x_{m_r} - \overline l}       | r = \size {x_{m_r} - l + l - \overline l}       | c =  }} {{eqn | o = \le       | r = \size {x_{m_r} - l} + \size {l - \overline l}       | c = Triangle Inequality }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2 = \epsilon       | c =  }} {{end-eqn}} Thus we have shown that: :$\forall r > R: \size {x_{m_r} - \overline l} < \epsilon$ So far, what has been shown is that, given any $\epsilon > 0$, there exists an infinite set of terms of $\sequence {x_n}$ which satisfy $\size {x_n - \overline l} < \epsilon$. Next it is shown how to construct a subsequence: :$\sequence {x_{n_r} }$ such that $x_{n_r} \to \overline l$ as $n \to \infty$. Take $\epsilon = 1$ in the above. Then: :$\exists n_1: \size {x_{n_1} - \overline l} < 1$ Now take $\epsilon = \dfrac 1 2$ in the above. Then: :$\exists n_2 > n_1: \size {x_{n_2} - \overline l} < \dfrac 1 2$ In this way a subsequence is contructed: :$\sequence {x_{n_r} }$ satisfying $\size {x_{n_r} - \overline l} < \dfrac 1 r$ But $\dfrac 1 r \to 0$ as $r \to \infty$ from the Sequence of Reciprocals is Null Sequence. From the Squeeze Theorem for Real Sequences, it follows that: :$\size {x_{n_r} - \overline l} \to 0$ as $r \to \infty$ Thus $\overline l \in L$ as required. {{qed|lemma}} A similar argument shows that the infimum $\underline l$ of $L$ is also in $L$. {{qed}}
Let $\sequence {x_n}$ be a sequence in $\R$ that converges to the limit $l \in \R$. Let $\epsilon > 0$.  Then also $\dfrac \epsilon 2 > 0$. Because $\sequence {x_n}$ converges to $l$, we have: :$\exists N: \forall n > N: \size {x_n - l} < \dfrac \epsilon 2$ So if $m > N$ and $n > N$, then: {{begin-eqn}} {{eqn | l = \size {x_n - x_m}       | r = \size {x_n - l + l - x_m} }} {{eqn | o = \le       | r = \size {x_n - l} + \size {l - x_m}       | c = Triangle Inequality }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = by choice of $N$ }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {x_n}$ is a Cauchy sequence. {{qed}}
From Real Number Line is Complete Metric Space, $\R$ under the usual metric is a metric space. The result then follows as a special case of Convergent Sequence in Metric Space is Cauchy Sequence. {{qed}}
From Sum of Geometric Sequence: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^{n - 1} x^j       | r = \frac {x^n - 1} {x - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\dfrac a b}^n - 1       | r = \paren {\dfrac a b - 1} \sum_{j \mathop = 0}^{n - 1} \paren {\dfrac a b}^j       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\dfrac {a^n - b^n} {b^n} }       | r = \paren {\dfrac {a - b} b} \paren {\paren {\dfrac a b}^{n - 1} + \paren {\dfrac a b}^{n - 2} + \dotsb + \paren {\dfrac a b}^1 + 1}       | c =  }} {{eqn | ll= \leadsto       | l = a^n - b^n       | r = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + \dotsb + a b^{n - 2} + b^{n - 1} }       | c = multiplying both sides by $b^n$ }} {{end-eqn}} {{qed}}
The proof will proceed by the Principle of Complete Finite Induction on $\Z_{>0}$. Let $S$ be the set defined as: :$\displaystyle S := \set {n \in \Z_{>0}: a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j}$ That is, $S$ is to be the set of all $n$ such that: :$\displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$ === Basis for the Induction === We have that: :$\displaystyle a^1 - b^1 = \paren {a - b} \sum_{j \mathop = 0}^0 a^{1 - 0 - 1} b^j$ So $1 \in S$. This is the basis for the induction. === Induction Hypothesis === It is to be shown that if $r \in S$ for all $r$ such that $1 \le r \le k$, then it follows that $k + 1 \in S$. This is the induction hypothesis: :$\forall r \in \Z_{>0}: 1 \le r \le k: \displaystyle a^r - b^r = \paren {a - b} \sum_{j \mathop = 0}^{r - 1} a^{r - j - 1} b^j$ It is to be demonstrated that it follows that: :$\displaystyle a^{k + 1} - b^{k + 1} = \paren {a - b} \sum_{j \mathop = 0}^k a^{k - j} b^j$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = a^{k + 1} - b^{k + 1}       | r = \paren {a + b} \paren {a^k - b^k} - a b \paren {a^{k - 1} - b^{k - 1} }       | c =  }} {{eqn | r = \paren {a + b} \paren {a - b} \sum_{j \mathop = 0}^{k - 1} a^{k - j - 1} b^j - a b \paren {a - b} \sum_{j \mathop = 0}^{k - 2} a^{k - j - 2} b^j       | c = Induction Hypothesis }} {{eqn | r = \paren {a - b} \paren {\paren {a + b} \sum_{j \mathop = 0}^{k - 1} a^{k - j - 1} b^j - a b \sum_{j \mathop = 0}^{k - 2} a^{k - j - 2} b^j}       | c =  }} {{eqn | r = \paren {a - b} \paren {\sum_{j \mathop = 0}^{k - 1} a^{k - j} b^j + \sum_{j \mathop = 0}^{k - 1} a^{k - j - 1} b^{j + 1} - \sum_{j \mathop = 0}^{k - 2} a^{k - j - 1} b^{j + 1} }       | c =  }} {{eqn | r = \paren {a - b} \paren {\sum_{j \mathop = 0}^{k - 1} a^{k - j} b^j + a^0 b^k} }} {{eqn | r = \paren {a - b} \sum_{j \mathop = 0}^k a^{k - j} b^j }} {{end-eqn}} So $\forall r \in S: 0 \le r \le k: r \in S \implies k + 1 \in S$ and the result follows by the Principle of Complete Finite Induction: :$\forall n \in \Z_{>0}: \displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$ {{qed}}
Let $P = \paren {x - \xi} Q$. Then: :$\map P \xi = \map Q \xi \cdot 0 = 0$ Conversely, let $\map P \xi = 0$. By the Division Theorem for Polynomial Forms over Field, there exist polynomials $Q$ and $R$ such that: :$P = \map Q {x - \xi} + R$ and: :$\map \deg R < \map \deg {x - \xi} = 1$ Evaluating at $\xi$ we have: :$0 = \map P \xi = \map R \xi$ But: :$\deg R = 0$ so: :$R \in K$ In particular: :$R = 0$ Thus: :$P = \map Q {x - \xi}$ as required. The fact that $\map \deg Q = n - 1$ follows from: :Ring of Polynomial Forms is Integral Domain and: :Degree of Product of Polynomials over Integral Domain. We can then apply this result to: :$\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$ We can progressively work through: :$\map P x = \paren {x - \xi_1} \map {Q_{n - 1} } x$  where $\map {Q_{n - 1} } x$ is a polynomial of order $n - 1$. Then, substituting $\xi_2$ for $x$: :$0 = \map P {\xi_2} = \paren {\xi_2 - \xi_1} \map {Q_{n - 1} } x$ Since $\xi_2 \ne \xi_1$: :$\map {Q_{n - 1} } {\xi_2} = 0$ and we can apply the above result again: :$\map {Q_{n - 1} } x = \paren {x - \xi_2} \map {Q_{n - 2} } x$ Thus: :$\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \map {Q_{n - 2} } x$ and we then move on to consider $\xi_3$. Eventually we reach: :$\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \dotsm \paren {x - \xi_n} \map {Q_0} x$ $\map {Q_0} x$ is a polynomial of zero degree, that is a constant polynomial. The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \displaystyle \sum_{k \mathop = 1}^n a_k       | r = \sum_{k \mathop = 1}^n \left({b_k - b_{k + 1} }\right)       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 1}^n b_{k + 1}       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 2}^{n + 1} b_k       | c = Translation of Index Variable of Summation }} {{eqn | r = b_1 + \sum_{k \mathop = 2}^n b_k - \sum_{k \mathop = 2}^n b_k - b_{n + 1}       | c =  }} {{eqn | r = b_1 - b_{n + 1}       | c =  }} {{end-eqn}} If $\left \langle {b_k} \right \rangle$ converges to zero, then $b_{n + 1} \to 0$ as $n \to \infty$. Thus: : $\displaystyle \lim_{n \mathop \to \infty} s_n = b_1 - 0 = b_1$ So: : $\displaystyle \sum_{k \mathop = 1}^\infty a_k = b_1$ {{Qed}}
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^N \paren {\lambda a_n + \mu b_n}       | r = \lambda \sum_{n \mathop = 1}^N a_n + \mu \sum_{n \mathop = 1}^N b_n       | c = Linear Combination of Indexed Summations }} {{eqn | o = \to       | r = \lambda \alpha + \mu \beta \text{ as } N \to \infty       | c = Combination Theorem for Sequences }} {{end-eqn}} {{qed}}
Let $\epsilon > 0$. Let $\displaystyle \sum_{k \mathop = 0}^\infty a_k$ converge to $s$. Then its sequence of partial sums $\sequence {s_N}$, where $\displaystyle s_N = \sum_{n \mathop = 1}^N a_n$, is a Cauchy sequence. So: : $\displaystyle \exists N: \forall k, m: k \ge m \ge N: \size {\sum_{l \mathop = m}^k a_l} < \frac \epsilon 3$ From Abel's Lemma: Formulation 2, we have: :$\displaystyle \sum_{k \mathop = m}^n u_k v_k = \sum_{k \mathop = m}^{n-1} \paren {\paren {\sum_{l \mathop = m}^k u_l} \paren {v_k - v_{k+1}} } + v_n \sum_{k \mathop = m}^n u_k$ We apply this, with $u_k = a_k$ and $v_k = x^k$: :$\displaystyle \sum_{k \mathop = m}^n a_k x^k = \sum_{k \mathop = m}^{n-1} \paren {\paren {\sum_{l \mathop = m}^k a_l} \paren {x^k - x^{k+1}} } + x^n \sum_{k \mathop = m}^n a_k$ So it follows that $\forall n \ge m \ge N$ and $\forall 0 < x < 1$, we have: {{begin-eqn}} {{eqn | l = \size {\sum_{k \mathop = m}^n a_k x^k}       | o = <       | r = \paren {1 - x} \sum_{k \mathop = m}^{n-1} \frac \epsilon 3 x^k + \frac \epsilon 3 x^n       | c = replacing instances of $\displaystyle \sum_{l \mathop = m}^k a_l$ with $\dfrac \epsilon 3$ }} {{eqn | o = <       | r = \frac \epsilon 3 \paren {1 - x} \frac {1 - x^n} {1 - x} + \frac \epsilon 3 x^n       | c = Sum of Geometric Progression }} {{eqn | r = \frac \epsilon 3       | c =  }} {{end-eqn}} So we conclude that: : $\displaystyle \size {\sum_{k \mathop = N}^\infty a_k x^k} \le \frac \epsilon 3$ Next, note that from the above, we have $\forall x: 0 < x < 1$: :$\displaystyle \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} \le \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} + \frac \epsilon 3 + \frac \epsilon 3$ But for finite $n$, we have that $1 - x^n \to 0$ as $x \to 1^-$. Thus: :$\displaystyle \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} \to 0$ as $x \to 1^-$ So: : $\displaystyle \exists \delta > 0: \forall x: 1 - \delta < x < 1: \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} < \frac \epsilon 3$ So, for any given $\epsilon > 0$, we can find a $\delta > 0$ such that, for any $x$ such that $1 - \delta < x < 1$, it follows that: : $\displaystyle \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} < \epsilon$ That coincides with the definition for the limit from the left. The result follows. {{qed}}
Follows directly from Euler's Formula $e^{i z} = \cos z + i \sin z$, by plugging in $z = \pi$: :$e^{i \pi} + 1 = \cos \pi + i \sin \pi + 1 = -1 + i \times 0 + 1 = 0$ {{qed}}
This theorem is a restatement of Image of Interval by Continuous Function is Interval. From Image of Interval by Continuous Function is Interval, the image of $\openint a b$ under $f$ is also a real interval (but not necessarily open). Thus if $k$ lies between $\map f a$ and $\map f b$, it must be the case that: :$k \in \Img {\openint a b}$ The result follows. {{qed}}
Let $\sequence {s_n}$ be the sequence of partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$. Let $\sequence {s'_n}$ be the sequence of partial sums of $\displaystyle \sum_{n \mathop = N}^\infty a_n$. It will be shown that $\sequence {s'_n}$ fulfils the Cauchy criterion. That is: :$\forall \epsilon \in \R_{>0}: \exists N: \forall m, n > N: \size {s'_n - s'_m} < \epsilon$ Let $\epsilon \in \R_{>0}$ be a strictly positive real number. As $\sequence {s_n}$ is convergent, it conforms to the Cauchy criterion by Convergent Sequence is Cauchy Sequence. Thus: :$\exists N: \forall m, n > N: \size {s_n - s_m} < \epsilon$ Now: {{begin-eqn}} {{eqn | l = s_n       | r = \sum_{k \mathop = 1}^n a_k       | c =   }} {{eqn | r = \sum_{k \mathop = 1}^{N - 1} a_k + \sum_{k \mathop = N}^n a_k        | c = Indexed Summation over Adjacent Intervals }} {{eqn | r = s_{N - 1} + s'_n }} {{end-eqn}} and similarly: :$s_m = s_{N - 1} + s'_m$ Thus: :$s'_n = s_n - s_{N - 1}$ and: :$s'_m = s_m - s_{N - 1}$ So: {{begin-eqn}} {{eqn | l = \size {s_n - s_m}       | o = <       | r = \epsilon       | c =  }} {{eqn | ll= \leadsto       | l = \size {s_n - s_{N - 1} - s_m + s_{N - 1} }       | o = <       | r = \epsilon       | c = }} {{eqn | ll= \leadsto       | l = \size {\paren {s_n - s_{N - 1} } - \paren {s_m - s_{N - 1} } }       | o = <       | r = \epsilon       | c =  }} {{eqn | ll= \leadsto       | l = \size {\size {s_n - s_{N - 1} } - \size {s_m - s_{N - 1} } }       | o = <       | r = \epsilon       | c = Triangle Inequality }} {{eqn | ll= \leadsto       | l = \size {s'_n - s'_m}       | o = <       | r = \epsilon       | c =  }} {{end-eqn}} So $\displaystyle \sum_{n \mathop = N}^\infty a_n$ fulfils the Cauchy criterion. By Convergent Sequence is Cauchy Sequence it follows that it is convergent. Now it is shown that $\displaystyle \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$. We have that $\sequence {s_n}$ is convergent, Let its limit be $l$. Thus we have: :$\displaystyle l = \sum_{n \mathop = 1}^\infty a_n = s_{N - 1} + \sum_{n \mathop = N}^\infty a_n$ So: :$\displaystyle \sum_{n \mathop = N}^\infty a_n = l - s_{N - 1}$ But $s_{N - 1} \to l$ as $N - 1 \to \infty$. The result follows. {{qed}}
Let $\epsilon > 0$. As $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ converges, its tail tends to zero. So: :$\displaystyle \exists N: \forall n > N: \sum_{k \mathop = n + 1}^\infty b_k < \epsilon$ Let $\sequence {a_n}$ be the sequence of partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$. Then $\forall n > m > N$: {{begin-eqn}} {{eqn | l = \cmod {a_n - a_m}       | r = \cmod {\paren {a_1 + a_2 + \cdots + a_n} - \paren {a_1 + a_2 + \cdots + a_m} }       | c =  }} {{eqn | r = \cmod {a_{m + 1} + a_{m + 2} + \cdots + a_n}       | c = Indexed Summation over Adjacent Intervals }} {{eqn | o = \le       | r = \cmod {a_{m + 1} } + \cmod {a_{m + 2} } + \cdots + \cmod {a_n}       | c = Triangle Inequality for Indexed Summations }} {{eqn | o = \le       | r = b_{m + 1} + b_{m + 2} + \cdots + b_n       | c =  }} {{eqn | o = \le       | r = \sum_{k \mathop = n + 1}^\infty b_k }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} So $\sequence {a_n}$ is a Cauchy sequence. The result follows from Real Number Line is Complete Metric Space or Complex Plane is Complete Metric Space. {{qed}}
From the statement of the theorem, it is necessary that $\forall n: a_n \ne 0$; otherwise $\size {\dfrac {a_{n + 1} } {a_n} }$ is not defined. Here, $\size {\dfrac {a_{n + 1} } {a_n} }$ denotes either the absolute value of $\dfrac {a_{n + 1} } {a_n}$, or the complex modulus of $\dfrac {a_{n + 1} } {a_n}$. === Absolute Convergence === Suppose $l < 1$. Let us take $\epsilon > 0$ such that $l + \epsilon < 1$. Then: :$\exists N: \forall n > N: \size {\dfrac {a_n} {a_{n - 1} } } < l + \epsilon$ Thus: {{begin-eqn}} {{eqn | l = \size {a_n}       | r = \size {\frac {a_n} {a_{n - 1} } } \size {\frac {a_{n - 1} } {a_{n - 2} } } \dotsm \size {\frac {a_{N + 2} } {a_{N + 1} } } \size {a_{N + 1} }       | c =  }} {{eqn | o = <       | r = \paren {l + \epsilon}^{n - N - 1} \size {a_{N + 1} }       | c =  }} {{end-eqn}} By Sum of Infinite Geometric Progression, $\displaystyle \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ converges. So by the the corollary to the comparison test, it follows that $\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n}$ converges absolutely too. {{qed}} === Divergence === Suppose $l > 1$. Let us take $\epsilon > 0$ small enough that $l - \epsilon > 1$. Then, for a sufficiently large $N$, we have: {{begin-eqn}} {{eqn | l = \size {a_n}       | r = \size {\frac {a_n} {a_{n - 1} } } \size {\frac {a_{n - 1} } {a_{n - 2} } } \dotsm \size {\frac {a_{N + 2} } {a_{N + 1} } } \size {a_{N + 1} }       | c =  }} {{eqn | o = >       | r = \paren {l - \epsilon}^{n - N + 1} \size {a_{N + 1} }       | c =  }} {{end-eqn}} But $\paren {l - \epsilon}^{n - N + 1} \size {a_{N + 1} } \to \infty$ as $n \to \infty$. So $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ diverges. {{qed}}
Let us pick $N \in \N$. Then $\exists n_1 > N: x_{n_1} \ge b$. Again, $\exists n_2 > n_1: x_{n_2} \ge b$. And so on: for each $n_k$ we find, $\exists n_{k+1} > n_k: x_{n_{k+1}} \ge b$. In this way we can build a subsequence of $\sequence {x_n}$ each of whose terms are $b$ or bigger. By the Bolzano-Weierstrass Theorem, this subsequence itself contains a subsequence $\sequence {x_{n_r} }$ which is convergent. Now, suppose $x_{n_r} \to l$ as $r \to \infty$. Since $x_{n_r} \ge b$ it follows from Lower and Upper Bounds for Sequences that $l \ge b$. {{qed}}
=== Upper Bound === First we show that: :$\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$ {{AimForCont}} this proposition were to be false. That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \ge \overline l + \epsilon$. From Limit of Subsequence of Bounded Sequence it would follow that there exists a convergent subsequence whose limit was $l \ge \overline l + \epsilon$. This would contradict the definition of $\overline l$. {{qed|lemma}} === Lower Bound === Next, in the same way, we show that: : $\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$ {{AimForCont}} this proposition were to be false. That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \le \underline l + \epsilon$. From Limit of Subsequence of Bounded Sequence it would follow that there exists a convergent subsequence whose limit was $l \le \underline l + \epsilon$. This would contradict the definition of $\underline l$. {{qed}}
=== 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 $n \to \infty$. {{qed|lemma}} === 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. {{qed}}