Limit

Finite Limits

• Definitions: $(a_n)$ converges to $L$
  • $\underset{n\to \infty }{\lim }{a}_n=L$
  • The sequence $(a_n)$ converges to (or tends to) the limit $L$
  • For each real number $\epsilon >0$, there exists a natural number $N$ such that, for every natural number $n>N$, we have $|a_n-L|<\epsilon$
  • $\forall \epsilon >0,\exists N\in \mathbb{N}:\forall n\in \mathbb{N}\left(n\ge N⟹|a_n-L|<\epsilon \right)$
• Definitions: Convergent Sequence
  • $(a_n)$ is a convergent sequence
  • $\exists x\in \mathbb{R}:\underset{n\to \infty }{\lim }{a}_n=x$
  • $\exists x\in \mathbb{R}:\forall \epsilon >0,\exists N\in \mathbb{N}:\forall n\in \mathbb{N}\left(n\ge N⟹|a_n-x|<\epsilon \right)$
  • not-in-course (3.36. Cauchy’s convergence test) $\forall \epsilon >0\exists N\in \mathbb{N}:m,n>N⟹|a_n-a_m|<\epsilon .$ (in words: $(a_n)$ is Cauchy sequence )
  • (3.34todo) $(a_n)$ is bounded and has only one subsequential limit
  • (q3.56) $(a_n)$ is bounded and $\underset{n\to \infty }{\mathrm{liminf}}{a}_n=\underset{n\to \infty }{\mathrm{limsup}}{a}_n$

Arithmetic

Assuming $(a_n)$ and $(b_n)$ are convergent (2.28)

Sequence Limit Rules
Multiple Rule	$\underset{n\to \infty }{\lim }(c{a}_n)=c\cdot \underset{n\to \infty }{\lim }{a}_n$
Sum Rule	$\underset{n\to \infty }{\lim }(a_n\pm b_n)=\underset{n\to \infty }{\lim }{a}_n\pm \underset{n\to \infty }{\lim }{b}_n$
Product Rule	$\underset{n\to \infty }{\lim }(a_n\cdot b_n)=\left(\underset{n\to \infty }{\lim }{a}_n\right)\cdot \left(\underset{n\to \infty }{\lim }{b}_n\right)$
Quotient Rule	$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\frac{\underset{n\to \infty }{\lim }{a}_n}{\underset{n\to \infty }{\lim }{b}_n}$	(provided $\underset{n\to \infty }{\lim }{b}_n\ne 0$)
---- Reciprocal Rule	$\underset{n\to \infty }{\lim }\left(\frac{1}{a_n}\right)=\frac{1}{\underset{n\to \infty }{\lim }{a}_n}$
	$\underset{n\to \infty }{\lim }{a}_n^k={\left(\underset{n\to \infty }{\lim }{a}_n\right)}^k$	$\forall k\in \mathbb{N}$
	$(\forall n,a_n>0)⟹\underset{n\to \infty }{\lim }\sqrt{a_n}=\sqrt{\underset{n\to \infty }{\lim }{a}_n}$

Theorems

• (2.29) Shift Rule Let $N$ be a natural number. Let $(a_n)$ be a sequence. Then $a_n\to a$ if and only if the shifted sequence $a_{N+n}\to a$.

• Let $(a_n)$ be a convergent sequence

  • (2.12) Limit Uniqueness $((a_n)\underset{n\to \infty }{⟶}L\wedge (a_n)\underset{n\to \infty }{⟶}M)⟹L=M$
  • (2.16) $(a_n)$ is bounded
  • (2.17) A sequence $(b_n)$ that obtained by changing a finite number of terms of $(a_n)$, has the same limit as $(a_n)$
  • (2.25) if $\forall n,a_n\ne 0$ and $\underset{n\to \infty }{\lim }{a}_n\ne 0$, then $(1/a_n)$ is bounded
  • (2.26) if $\underset{n\to \infty }{\lim }{a}_n\ne 0$ then for almost all $n$, we have $a_n\ne 0$
  • (q3.73) $\underset{n\to \infty }{\lim }(a_{n+k}-a_n)=0$ (for all $k\in \mathbb{N}$)

• Let $(a_n)$ and $(b_n)$ be two convergent sequences:

  • (2.30) if $\underset{n\to \infty }{\lim }{a}_n<\underset{n\to \infty }{\lim }{b}_n$ then, for almost all $n$, we have $a_n<b_n$
  • (2.31) if $a_n\le b_n$ for each $n$, then $\underset{n\to \infty }{\lim }{a}_n\le \underset{n\to \infty }{\lim }{b}_n$
  • (2.32) Squeeze theorem - Let $(x_n)$ be a sequence, where $a_n\le x_n\le b_n$ for almost all $n$, then, $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{b}_n=L⟹\underset{n\to \infty }{\lim }{x}_n=L$

Null Sequence

• (d2.21) $(a_n)$ is null sequence

• $\underset{n\to \infty }{\lim }{a}_n=0$

• (q2.20a) $(|a_n|)$ is null sequence

• $\forall \epsilon >0,\exists N\in \mathbb{N}:\forall n\in \mathbb{N},(n\ge N⟹|a_n|<\epsilon )$

• A null sequence is bounded sequence

• (2.22) A product of null and bounded sequences is null sequence

• Squeeze theorem - if $(b_n)$ is null sequence and $|a_n|<b_n$ for all $n$, then, $(a_n)$ is null sequence

Strategy for using the definition of null sequence

• To show that $(a_n)$ is null, solve the inequality $|a_n|<\epsilon$ to find a number $N$ (generally depending on $\epsilon$) such that $|a_n|<\epsilon$ for all $n>N$.
• To show that $(a_n)$ is not null, find ONE value of $\epsilon$ for which there is NO number $N$ such that $|a_n|<\epsilon$, for all $n>N$.

Infinite Limits

• Infinity
  • $\underset{n\to \infty }{\lim }{a}_n=\infty$
  • $a_n\to \infty$
  • $(a_n)$ is tend to infinity
  • For every real number $K$, there is a natural number $N$ such that for every natural number $n\ge N$, we have  $a_n>K$; that is, the sequence terms are eventually larger than any fixed $K$.
  • $\forall K\in \mathbb{R}\left(\exists N\in \mathbb{N}\left(\forall n\in \mathbb{N}\left(n\ge N⟹a_n>K\right)\right)\right)$
  • (2.39) $\underset{n\to \infty }{\lim }(-a_n)=-\infty$
• Minus Infinity
  • $\underset{n\to \infty }{\lim }{a}_n=-\infty$
  • $a_n\to -\infty$
  • $(a_n)$ is tend to minus infinity
  • $\forall K\in \mathbb{R}\left(\exists N\in \mathbb{N}\left(\forall n\in \mathbb{N}\left(n\ge N⟹a_n<K\right)\right)\right)$

Theorems

• If a sequence tends to infinity or minus infinity, then:

  • (2.40) it is unbounded
  • (2.41 it is divergent

• (2.44)todo like 2.29 but for infinite limit

• Squeeze Theorem for infinite limit

  • (2.45) if $a_n\to \infty$ and $b_n\ge a_n$ for almost all $n$, then $b_n\to \infty$
  • (q2.40) if $a_n\to -\infty$ and $b_n\le a_n$ for almost all $n$, then $b_n\to -\infty$

Arithmetics

• (2.43, q2.39) arithmetics of infinite limits

$a_n$	$b_n$	$⟹$	$a_n+b_n$	$a_n{b}_n$	$a_n/b_n$
$\infty$	$\infty$		$\infty$	$\infty$
$-\infty$	$-\infty$		$-\infty$	$\infty$
$-\infty$	$\infty$			$-\infty$
$\infty$	$L\in \mathbb{R}$		$\infty$
$\infty$	$L>0$			$\infty$
$0$	$\infty$		$\infty$		$0$

• $a_n\to \infty ⟹1/a_n\to 0$

• (q2.57a) $|a_n|\to \infty ⟹1/a_n\to 0$

• $a_n\to 0\wedge a_n>0⟹1/a_n\to \infty$

• $a_n\to 0\wedge a_n<0⟹1/a_n\to (-\infty )$

• $a_n\to \infty$ and $b_n$ is bounded $⟹a_n+b_n\to \infty$

• $\frac{a_n}{b_n}\to 1\wedge a_n\to 0⟹b_n\to 0$

Examples

• (2.37) $\underset{n\to \infty }{\lim }n=\infty$

• (e2.11) $\underset{n\to \infty }{\lim }\frac{n}{n+1}=1$

• (e2.12) $\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n}}=0$

• $\underset{n\to \infty }{\lim }\sqrt[n]{n}=1$

• $c\in \mathbb{R}$

  • $\underset{n\to \infty }{\lim }\frac{c^n}{n!}=0$

• $-1<c<1\equiv |c|<1$

  • (q2.54) $\underset{n\to \infty }{\lim }(1+c+\cdots +c^n)=\frac{1}{1+c}$
  • (2.33) $\underset{n\to \infty }{\lim }{c}^n=0$

• $c>0$

  • (2.34) $\underset{n\to \infty }{\lim }\sqrt[n]{c}=1$
  • $\underset{n\to \infty }{\lim }\frac{n^c}{n!}=0$
  • $\underset{n\to \infty }{\lim }\frac{1}{n^c}=0$
    • (2.10) $\underset{n\to \infty }{\lim }\frac{1}{n}=0$

• $c>1$

  • (q2.41) $\underset{n\to \infty }{\lim }{c}^n=\infty$

• $|c|>1$

  • $\underset{n\to \infty }{\lim }\frac{1}{c^n}=0$ (by 2.33,q2.20)

• (q3.20) $\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)=1$

• (6.19) $\underset{n\to \infty }{\lim }{\left(\frac{n+c}{n}\right)}^n=e^c$

  • (special case c=1) $\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=\underset{n\to \infty }{\lim }{\left(\frac{n+1}{n}\right)}^n=e$
  • (q3.20d, special case c=2) $\underset{n\to \infty }{\lim }{\left(1+\frac{2}{n}\right)}^n=\underset{n\to \infty }{\lim }{\left(\frac{n+2}{n}\right)}^n=e^2$
  • (q3.20b) $\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n+1}\right)}^n=e$
  • (q3.20a) $\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^{n+1}=e$

• (6.15) $\underset{n\to \infty }{\lim }\left(a_n^{b_n}\right)={\left(\underset{n\to \infty }{\lim }{a}_n\right)}^{\left(\underset{n\to \infty }{\lim }{b}_n\right)}$ (assuming $\lim a_n>0$, and $\lim b_n\in \mathbb{R}$)

• (2.49) assuming $k\in \mathbb{N}$, and $1<r\in \mathbb{R}$

  • $\underset{n\to \infty }{\lim }\left(\frac{n^k}{r^n}\right)=0$

• (6.4) if $(q_n{)}_{n=1}^{\infty }$ is a sequence of rationals that tends to 0, and $a>0$, then $\underset{n\to \infty }{\lim }{a}^{q_n}=1$

• (q2.29) Let $a_1,\dots ,a_k\ge 0$, then $\underset{n\to \infty }{\lim }\sqrt[n]{\sum _{i=1}^k{a}_i^n}=\max \{a_1,\dots ,a_k\}$

• todo For any continuous function $f$, if ${\lim }_{n\to \infty }{a}_n$ exists, then ${\lim }_{n\to \infty }f\left(a_n\right)$ exists too.

• (some exam) given $(a_n)$ is positive

  • $\underset{n\to \infty }{\lim }(a_n)>0⟹\underset{n\to \infty }{\lim }\sqrt{a_n}=1$
  • $\underset{n\to \infty }{\lim }(a_n)=\infty ⟹\underset{n\to \infty }{\lim }\sqrt{a_n}>1$
  • $\underset{n\to \infty }{\lim }(a_n)=\infty ⟹\sqrt{a_n}>1$ for almost all $n$

Ratio test

• $(a_n)$ is sequence (all are nonzero)
  • (2.47) (given $0<r<1$) if $\left|\frac{a_{n+1}}{a_n}\right|\le r$ for almost all $n$ then $\underset{n\to \infty }{\lim }(a_n)=0$
  • (2.48) $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|<1⟹\underset{n\to \infty }{\lim }(a_n)=0$
    • if $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}>1$ (or $=\infty$) and $a_n>0$, then $\underset{n\to \infty }{\lim }(a_n)=\infty$

Cantor’s Lemma

• (3.22) Cantor’s intersection theorem (Cantor’s Lemma, Nested Intervals Theorem (Property))
  • A sequence of nested intervals is a sequence $(I_n{)}_{n=1}^{\infty }$ of non-empty, closed intervals, $I_n=[a_n,b_n]$, satisfying:
    • $I_1\supseteq I_2\supseteq I_3\supseteq \dots$
    • $\underset{n\to \infty }{\lim }(b_n-a_n)=0$
  • The intersection of a sequence $(I_n{)}_{n=1}^{\infty }$ of nested intervals is ${\bigcap }_{n=1}^{\infty }{I}_n=\{x\}$, and $x=\underset{n\to \infty }{\lim }(a_n)=\underset{n\to \infty }{\lim }(b_n)$

Other form of Cantor’s intersection theorem: Let $(a_n{)}_{n=1}^{\infty }$ and $(b_n{)}_{n=1}^{\infty }$ be two sequence of real numbers that satisfy: $a_n\le a_{n+1}<b_{n+1}\le b_n$ for every $n\in \mathbb{N}$, and $\lim (b_n-a_n)=0$. Then there exists a real number $c$ such that $\lim a_n=\lim b_n=c$. The number $c$ is the unique real number that satisfies $a_n\le c\le b_n$