Real Sequences

This is following Goldberg’s old book on real analysis

definition: a sequence of real numbers is a function $f : \mathbb{N} \rightarrow {\mathbb R}$ . Classical notation $\{f_n\}$ . A subsequence is a function $f \circ g$ where $g$ is a strictly increasing sequence of positive integers.

lemma: A strictly increasing sequence of positive numbers $s$ is such that for all positive integers, $s_n \ge n$ .

the claim is true when $n = 1$ . Assume it to be true for $n-1$. Then $s_n \ge s_{n-1} + 1 \ge n-1 + 1 = n$ and so the induction hypothesis is verified.

definition: the limit of a real sequence $f$ is a real number $L$ such that for any $\epsilon > 0$ , there is an $N$ such that for all $n$ for which $n \ge N$ , $|f_n - L| < \epsilon$ . This is denoted by $\lim f = L$ or other variants.

lemma: if $s$ is a sequence of non-negative real numbers and $\lim s$ exists, then $\lim s$ is a non-negative real number.

If $\lim s < 0$ then there is an $N$ such that for all $n \ge N$ , $\lim s < \lim s - s_n < -\lim s$ . But then $s_n < 0$ , which is false.

definition: a sequence $s$ is convergent if $\lim s$ exists. Otherwise it is divergent.

lemma: a convergent sequence $s$ has one limit.

Let $L,M$ be limits of $s$ . Then $|L - M| < |L - s_n| + |s_n - M|$ , and for any $\epsilon > 0$ , there is an $N$ such that $|L - s_n | < \epsilon / 2, |M - s_n | < \epsilon /2$ , it follows that $|L - M| < \epsilon$ . Since this is true for all $\epsilon > 0$ , $L = M$ .

lemma: a subsequence of a convergent sequence $s$ is a convergent sequence with limit $\lim s$ .

Let $s$ be a sequence convergent to $L$ and $m$ be a strictly increasing sequence of positive integers. Then for any $\epsilon, |s \circ m(n) - L | \le \epsilon$ when $n \ge N$ where $N$ is an integer such that for all $n \ge N, |s(n) - L| < \epsilon$ . Hence $s \circ m$ is a convergent subsequence.

definition: a sequence $s$ of real numbers is divergent to infinity (convergent to infinity) if for any real number $M$ there is an integer $N$ such that for all $n \ge N$ , $s_n \ge M$ . Analogous definition for minus infinity.

definition: if a sequence $s$ is not convergent to a real number or to infinity or minus infinity, then it is oscillatory.

lemma: convergent sequences are bounded sequences.

Choose $N$ such that for all $n \ge N, |s_n - \lim s | < 1$ . Then $|s_n| \le \max(|s_i| : i = 1..N-1, |\lim s| + 1)$ .

definition: a sequence $s$ is non-decreasing | non-increasing if for all $n, s_{n+1} \ge s_n | s_{n+1} \le s_n$ A monotone sequence is a sequence that is non-decreasing or a sequence that is non-increasing.

lemma: a non-decreasing sequence that is bounded above is convergent.

Suppose $s_n \le M$ for all $n$ , Then the set $S = \{s_n : \in \mathbb{N} \}$ is a nonempty set of real numbers bounded above by a real number and so the least upper bound exists as a real number and furthermore, $\sup_n s_n \le M$ . Given $\epsilon > 0$ there is an $N$ such that $\sup S - \epsilon < s_N \le \sup S$ . Since $s$ is non-decreasing, for all $n \ge N$ , $|\sup S - s_n | < \epsilon$ . Hence $\sup S$ is the limit of the sequence latex $s$ .

lemma: a non-decreasing sequence that is not bounded above diverges to infinity.

Given any $M$ there is an $N$ such that $M < s_N$ . and so for all $n \ge N$ , $M < s_n$ .

A number of arithmetic operations can be performed on convergent sequences: sequential addition / multiplication / scalar multiplication

lemma: if $s,t$ are convergent sequences, $c$ is a real number then $s+t, cs,s\cdot t$ are convergent sequences with limits $\lim s + \lim t, c \lim s \lim s \cdot \lim t$ .

the proofs follow from key steps

$|s_n + t_n - \lim s - \lim t | \le |s_n - \lim s| + |t_n - \lim t|$

$|c s_n - c \lim s| = |c| |s_n - \lim s|$

$|s_n t_n - \lim s \lim t| = |s_n - \lim s | |t_n| + |\lim s| |t _n - \lim t|$

with the latter result following from boundedness of a convergent sequence.

lemma: $\{x^n\}$ converges to zero when $x \in (0,1)$ and diverges to infinity when $x > 1$ .

In the first case $\{ x^n\}$ is bounded below by zero and decreasing. So it is convergent to some $L \ge 0$ . For the sequence $\{x \cdot x_n \}$ , $L = \lim_n x x_n = x \lim_n x_n = x L$ , and so $L = 0$

In the second case, $\{ x^n \}$ is an unbounded increasing sequence and so is divergent to infinity: it is increasing since $x_{n+1} = x x_n > x_n > 1$ . If it is bounded, then it is convergent to some $M >1$ and so $M = \lim_n x x^n = x \lim_n x^n = x M$ , which implies $x = 1$ , false.

corollary: if $s,t$ are convergent sequences such that sequentially $s \le t$ the $\lim s \le \lim t$ .

this follows since $t - s$ is a non-negative convergent sequence and so $\lim (t - s) = \lim t - \lim s \ge 0$ , and so $\lim t \ge \lim s$ .

Operations on Divergent Sequences

While there are nice algebraic properties available for convergent sequences, the same cannot be said for divergent sequences. For example if $s$ is divergent to infinity, then $-s$ is divergent to negative infinity, but their sequential sum $s - s = 0$ which is costant and therefore convergent. The product of the oscillatory sequence $\{(-1)^n\}$ with itself is a constant sequence, which is convergent.

However, restriction of the type of divergence can result in some nice properties

lemma: if $s,t$ are divergent to infinity, and $c > 0$ , then $st,s+t, cs$ are divergent to infinity.

given $M > 0$ , choose $N$ such that for all $n \ge N, t_n > 1, s_n > M$ . Then $s_n + t_n > M$ , $s_n t_n > M$ and so $st,s+t$ are divergent to infinity.

lemma: if $s$ is divergent to infinity and $t$ is bounded, then $s + t$ is divergent to infinity.

Limit Superior and Limit Inferior

Given a sequence $s$ that is bounded above, then $r = \{ \sup_{n \ge m} s_n \}$ is a non-increasing sequence of real numbers. If $s$ is bounded below, then $r$ has a limit defined by $\limsup s$ . Otherwise, $r$ is divergent to minus infinity and one writes $\limsup s = -\infty$ . If $s$ is not bounded above then $r_n = \infty$ and one writes $\limsup s = \infty$ .

The limit inferior is defined analogously.

lemma: if $s$ is bounded above and has a subsequence $s \circ m$ that is bounded below by $A$ , then $\limsup s \ge A$ .

Since $s$ is bounded above, $\{ \sup_{n \ge m} s_m\}$ is a non-increasing sequence of real numbers. Since an infinite number of values of $s$ are bounded below by $A$ , it follows that $\sup_{n \ge m} s_m \ge A$ . Hence $\limsup s \ge A$ .

lemma: if $s$ is bounded above but has no subsequence that is bounded below, then $\limsup s = -\infty$ .

In this case, the supremums $\{\sup_{n \ge m} s_m \}$ is a non-increasing sequence of real numbers which is not bounded below, and therefore is divergent to negative infinity.

lemma: if $s$ is convergent then $\limsup s = \lim s$ .

a convergent sequence is bounded and so $\limsup s$ and $\liminf s$ exist as real numbers. For any $\epsilon > 0$ the sequence $s$ has a subsequence $s \circ m$ which is bounded below by $\lim s - \epsilon$ and so $\limsup s \ge \lim s - \epsilon$ . But since there is an $N$ such that $\lim_s + \epsilon > s_n$ for all $s \ge N, \lim s + \epsilon \ge \limsup s$ . Since $\epsilon > 0$ is arbitrary, it follows that $\limsup s = \lim s$ .

The limit inferior is defined analogously, and dual results can be obtained.

lemma: if $s$ is a sequence of real numbers then $\liminf s \le \limsup s$ .

When $s$ is bounded, $\inf_{n \ge m} s_n \le \sup_{n \ge m} s_n$ . Since limits of both sides exist, $\liminf s \le \limsup s$ .

If $s$ is not bounded, then either $\limsup = \infty$ or $\liminf = -\infty$ and the condition is satisifed.

lemma: if $s$ is a sequence of real numbers such that there is a real number $L$ such that $\liminf s = \limsup s = L$ , then $s$ is convergent to $L$ .

Since $|\sup_{m \ge n} s_m - L| < \epsilon$ implies $s_m < L + \epsilon$ for all $m \ge N$ , and $|\inf_{m \ge n} s_m - L| < \epsilon$ implies $L - \epsilon < s_m < \epsilon$ for all $m \ge N$ it follows that $|s_m - L| < \epsilon$ when $m \ge N$ .

lemma: if $s$ is a sequence such that $\liminf s = \limsup s = \infty | -\infty$ , then $\lim s = \infty | -\infty$

lemma: if $s,t$ are sequences such that $s \le t$ , then $\liminf s \le \liminf t$ and $\limsup s \le \limsup t$ .

The nice algebraic properties that held for limits do not generally hold for limit inferior/limit superior. For example, if $s_n = (-1)^n$ and $t_n = -s_n$ then $s_n + t_n = 0$ but $\limsup s = \limsup t = 1$ .

lemma: if $s,t$ are bounded sequences of real numbers then $\limsup (s+t) = \limsup s + \limsup t$ and $\liminf (s+t) = \liminf s + \liminf t$

since $s_m \le \sup_{m \ge n} s_m$ and $t_m \le \sup_{m \ge n} t_m$ when $m \ge n$ , so $s_m + t_m \le \sup_{m \ge n} + \sup_{m \ge n} t_m$ and so $\sup_{m \ge n} (s_m + t_m) \le \sup_{m \ge n} + \sup_{m \ge n} t_m$ and $\limsup (s+t) \le \limsup s + \limsup t$ .

lemma: if $s$ is a bounded sequence of numbers $\limsup s = M$ then for any $\epsilon > 0$ then $s_n < M + \epsilon$ for all but finitely many $n$ and $s_n > M - \epsilon$ for infinitely many $n$ .

Since $\inf_n \sup_{m \ge n} s_m < M + \epsilon$ there is an $n$ such that $\sup_{m \ge n} s_m < M + \epsilon$ , and so for all $m \ge n, s_m < M + \epsilon$ . Since $M -\epsilon$ is a lower bound, for all $n \sup_{m \ge n} s_m > M - \epsilon$ and so there is an $m \ge n$ such that $s_m > M - \epsilon$ .

lemma: any bounded sequence of real numbers contains a convergent subsequence.

let $M = \limsup s_n$ . and choose $n_1$ such that $s(n_1) > M - 1$ . Choose $n_2 > n_1$ such that $s(n_2) > M - 1/2$ . Continuing, choose $n_k > n_{k-1}$ such that $s(n_k) > M - 1/k$ . For any $\epsilon > 0$ there is an $N$ such that $1/N < \epsilon$ , and so $s(n_k) > M - 1/k > M - 1/N > M - \epsilon$ for all $k \ge N$ . But since there is an $N_2$ such that $s(n) < M + \epsilon$ for all $n \ge N_2$ it follows that $|s_n - M| < \epsilon$ when $n \ge \max(N_2,N)$ . Since this is true for any $\epsilon > 0$ it follows that $\{s(n_k)\}$ is convergent.

definition: a cauchy sequence is a sequence $s$ such that for any $\epsilon > 0$ there is an $N$ such that for all $n,m \ge N, |s_n - s_m| < \epsilon$ .

lemma: a cauchy sequence is bounded.

lemma: a cauchy sequence is a convergent sequence.

lemma: a nested sequence of intervals whose lengths converge to zero contains exactly one point.

Since $a_1 \le a_n \le b_n \le b_1$ the sequence of left/right endpoints is bounded. The sequence $\{a_n\}$ is non-decreasing and bounded above so has a limit $L$ . The sequence $\{b_n\}$ is non-increasing and bounded below so has a limit $M$ . Since $\lim (b_n - a_n) = 0$ , $M - L = 0$ and so $M = L$ . Let $x,y \in \cap [a_n,b_n]$ and assume $y \ge x$ . Then $y - x \le b_n - a_n$ which implies $y - x \le M - L = 0$ . And so $y = x$ .

Summability of Sequences

definition: A sequence $s$ of real numbers is $(C,1)$ summable to $L$ if the sequence of averages $\text{avg}(s)$ is convergent where $\text{avg}(s)_n = n^{-1} \sum{i=1..n} s_i$ .

note that $\text{avg}(s + c) = \text{avg}(s) + c$ .

lemma: a sequence $s$ convergent to $L$ is such that $\text{avg}(s)$ is convergent to $L$ .

wlog assume that $s$ is convergent to zero. Then $s$ is bounded by $M$ . Then for any $\epsilon > 0$ there is an $N$ such that for all $n \ge N$ , $|s_n| < \epsilon$ . But then $|a_n| \le \frac{1}{n} \sum_{i=1..n} |s_i| \le \frac{1}{n} ( MN + \epsilon (n - N) )$ and so $\lim_n |a_n| \le \epsilon$ . Since this property is true for all $\epsilon > 0$ it follows that $\lim_n |a_n| = 0$ , So $\lim_n a_n = 0$ .

lemma: if $s,t$ are $(C,1)$ summable to $L,M$ then $s-t,s+t$ are $(C,1)$ summable to $L-M,L+M$

definition: a sequence $s$ is $(C,2)$ summable to $L$ if $(n * s_n)/(n * 1)$ is convergent to $L$ .

lemma: if $s$ is $(C,1)$ summable, it is $(C,2)$ summable.

when $s$ Is $(C,1)$ summable to zero,

$\tau_n \frac{n(n+1)}{2} = \sum_{i=1..n} s_{i} (n+1-i) = \sum_{i=1..n} s_i \sum_{j=1..n} \mathbb{I}(j \ge i) = \sum_{j=1..n} \sum_{i=1..j} s_i = \sum_{j=1..n} j \sigma_j$

and so when $N$ is chosen such that $|\sigma_n| < \epsilon/2$ ,

$|\tau_n| \frac{n(n+1)}{2} \le \sum_{j=1..N-1} jM + \sum_{j=N..n} \epsilon/2$

$|\tau_n| \le \frac{(N-1)N M}{n(n+1)} + \epsilon/2$

Real Sequences

Operations on Divergent Sequences

Limit Superior and Limit Inferior

Summability of Sequences

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