Limits of sequences

## Limits of sequences
            ---
            ## Distance between real numbers
            Suppose $x,y\in\mathbb R$.
            Then $\lvert x-y \rvert$ is the *distance* between them.

Note $\lvert x-y \rvert = \lvert y-x \rvert$, as appropriate for a distance.

So "$\lvert x-y \rvert < 5$" means "$x$ and $y$ are within $5$ of each other".

cf. vectors: $\lVert\mathbf x -\mathbf y\rVert$ is the distance between two vectors, the length of the vector starting from $\mathbf y$ and going to $\mathbf x$.
            ---
            ## Limit of a sequence
            #### Definition
            A sequence $(a_n)\_{n=1}^\infty$ *converges* to *limit* $\ell\in\mathbb R$ if

for all $\varepsilon>0$,   $\exists k\in\mathbb N$ such that $n>k \implies$ $\lvert a_n-\ell \rvert < \varepsilon$.

If so, we say "$\displaystyle\lim\_{n\to\infty} a_n = \ell$" or "$a_n\to\ell$ as $n\to\infty$".

Informally, we think of a limit as "something that the terms in the sequence get closer and closer to".
            But we need a more precise understanding for this course.

#### Examples
            If $a_n = \frac{-1}n$, then $\displaystyle\lim\_{n\to\infty} a_n = 0$.

If $b_n = \frac{n+3}{5-2n}$, then $\displaystyle\lim\_{n\to\infty} b_n = \frac{-1}2$.
 
 ---
 ## Limit of a sequence
 #### Definition
 A sequence $(a_n)\_{n=1}^\infty$ *converges* to *limit* $\ell\in\mathbb R$ if

for all $\varepsilon>0$,   $\exists k\in\mathbb N$ such that $n>k \implies$ $\lvert a_n-\ell \rvert < \varepsilon$.

#### Or, in words,
 No matter how small we pick $\varepsilon>0$,

we can always go far enough (ie. $k$ far) down the sequence to ensure that

every term after that point (ie. after $k$),

is within $\varepsilon$ of $\ell$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 
 ---
 ## Proof of sequence limit: example 1
 #### Theorem
 If $a_n = \frac{-1}n$, then $\displaystyle\lim\_{n\to\infty} a_n = 0$.
 #### Definition
 A sequence $(a_n)\_{n=1}^\infty$ *converges* to *limit* $\ell\in\mathbb R$ if

#### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Proof
 Fix any $\varepsilon>0$.
 Let $k=\lceil\frac1\varepsilon\rceil$.
 If $n>k$, then

#### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Proof
 Fix any $\varepsilon>0$.
 Let $k=\max\\{10,\lceil\frac{11}{3\varepsilon}\rceil\\}$.
 If $n>k$, then

$\lvert b_n - \ell \rvert$ $= \lvert \frac{n+3}{5-2n}-\frac{-1}2 \rvert = \lvert\frac{(2n+6)+(5-2n)}{10-4n}\rvert=\lvert\frac{11}{10-4n}\rvert = \frac{11}{4n-10} < \frac{11}{3n}$ $<$ $\frac{11}{3k} \leqslant \frac{11}{3\lceil11/3\varepsilon\rceil} \leqslant \frac{11}{3(11/3\varepsilon)} =$ $\varepsilon$.   □
 
 ---
 ## Deriving proof of sequence limit: eg 1
 Theorem:
 If $a_n = \frac{-1}n$, then $\displaystyle\lim\_{n\to\infty} a_n = 0$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.

Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert a_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert a_n - \ell \rvert$ $= \lvert \frac{-1}n - 0 \rvert = \frac1n$, and I want this to be $<\varepsilon$.

I may require $n>k$, so that $\frac1n<\frac1k$, so $\lvert a_n - \ell \rvert<$ $\frac1k$, so now I just need $\frac1k\leqslant\varepsilon$.

First guess: $k = \frac1\varepsilon$.
 This gives $\frac1k\leqslant\varepsilon$ as I wanted, but probably $\frac1\varepsilon \notin \mathbb N$. So use $k = \lceil\frac1\varepsilon\rceil$.
 
 ---
 ## Deriving proof of sequence limit: eg 1
 Theorem:
 If $a_n = \frac{-1}n$, then $\displaystyle\lim\_{n\to\infty} a_n = 0$.
 #### Rough work
 Fix any $\varepsilon>0$.

Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert a_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert a_n - \ell \rvert$ $= \lvert \frac{-1}n - 0 \rvert = \frac1n$, and I want this to be $<\varepsilon$.

I may require $n>k$, so that $\frac1n<\frac1k$, so $\lvert a_n - \ell \rvert<$ $\frac1k$, so now I just need $\frac1k\leqslant\varepsilon$.

First guess: $k = \frac1\varepsilon$.
 This gives $\frac1k\leqslant\varepsilon$ as I wanted, but probably $\frac1\varepsilon \notin \mathbb N$. So use $k = \lceil\frac1\varepsilon\rceil$.
 #### Proof
 Fix any $\varepsilon>0$.
 Let $k=\lceil\frac1\varepsilon\rceil$.
 If $n>k$, then

$\lvert a_n - \ell \rvert$ $= \lvert \frac{-1}n - 0 \rvert = \frac1n$ $<$ $\frac 1k = \frac1{\lceil1/\varepsilon\rceil} \leqslant \frac1{1/\varepsilon} =$ $\varepsilon$.   □
 
 ---
 ## Deriving proof of sequence limit: eg 2
 Theorem:
 If $b_n = \frac{n+3}{5-2n}$, then $\displaystyle\lim\_{n\to\infty} b_n = \frac{-1}2$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert b_n-\ell \rvert < \varepsilon$.

Calculate $\lvert b_n - \ell \rvert$ $= \lvert \frac{n+3}{5-2n}-\frac{-1}2 \rvert = \lvert\frac{(2n+6)+(5-2n)}{10-4n}\rvert=\lvert\frac{11}{10-4n}\rvert$, and I want to simplify.

If I assume $4n>10$, then $\lvert\frac{11}{10-4n}\rvert=\frac{11}{4n-10}$.

If I assume $n>10$, then $\frac{11}{4n-10}<\frac{11}{3n}$,
 and I want this to be $<\varepsilon$.

I may require $n>t$, so that $\frac{11}{3n}<\frac{11}{3t}$, so $\lvert b_n - \ell \rvert<$ $\frac{11}{3t}$, so now I just need $\frac{11}{3t}\leqslant\varepsilon$.

If I assume $4n>10$, then $\lvert\frac{11}{10-4n}\rvert=\frac{11}{4n-10}$.

If I assume $n>10$, then $\frac{11}{4n-10}<\frac{11}{3n}$,
 and I want this to be $<\varepsilon$.

So I pick $k =$ $\max\\{ \frac{10}4, 10, \lceil\frac{11}{3\varepsilon}\rceil \\} =$ $\max\\{ 10, \lceil\frac{11}{3\varepsilon}\rceil \\}$.
 
 ---
 ## Deriving proof of sequence limit: eg 2
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert b_n-\ell \rvert < \varepsilon$.

If I assume $4n>10$, then $\lvert\frac{11}{10-4n}\rvert=\frac{11}{4n-10}$.

If I assume $n>10$, then $\frac{11}{4n-10}<\frac{11}{3n}$,
 and I want this to be $<\varepsilon$.

First guess: $t = \frac{11}{3\varepsilon}$.
 This gives $\frac{11}{3t}\leqslant\varepsilon$ as I wanted, but probably $\frac{11}{3\varepsilon} \notin \mathbb N$. So use $t = \lceil\frac{11}{3\varepsilon}\rceil$.
 #### Proof
 Fix any $\varepsilon>0$.
 Let $k=\max\\{10,\lceil\frac{11}{3\varepsilon}\rceil\\}$.
 If $n>k$, then
 $\lvert b_n - \ell \rvert$ $= \lvert \frac{n+3}{5-2n}-\frac{-1}2 \rvert = \lvert\frac{(2n+6)+(5-2n)}{10-4n}\rvert=\lvert\frac{11}{10-4n}\rvert = \frac{11}{4n-10} < \frac{11}{3n}$ $<$ $\frac{11}{3k} \leqslant \frac{11}{3\lceil11/3\varepsilon\rceil} \leqslant \frac{11}{3(11/3\varepsilon)} =$ $\varepsilon$.   □
 
 ---
 ## Deriving proof of sequence limit: your turn 1
 Theorem:
 If $c_n = 1+\frac{(-1)^n}{6 n}$, then $\displaystyle\lim\_{n\to\infty} c_n = 1$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.

Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert c_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert c_n - \ell \rvert$ $=?$

…So pick $k=?$
 
 ---
 ## Deriving proof of sequence limit: your turn 1
 Theorem:
 If $c_n = 1+\frac{(-1)^n}{6 n}$, then $\displaystyle\lim\_{n\to\infty} c_n = 1$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert c_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert c_n - \ell \rvert$ $=\lvert 1+\frac{(-1)^n}{6 n}-1 \rvert = \frac{1}{6 n}$, and I want this to be $<\varepsilon$.

I may require $n>k$, so that $\frac1n<\frac1k$, so $\lvert c_n - \ell \rvert<$ $\frac1{6 k}$, so now I just need $\frac1{6 k}\leqslant\varepsilon$.

First guess: $k = \frac{1}{6\varepsilon}$.
 This gives $\frac1{6 k}\leqslant\varepsilon$ as required, but probably $\frac{1}{6\varepsilon} \notin \mathbb N$.
 So pick $k = \lceil\frac{1}{6\varepsilon}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: your turn 1
 Theorem:
 If $c_n = 1+\frac{(-1)^n}{6 n}$, then $\displaystyle\lim\_{n\to\infty} c_n = 1$.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert c_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert c_n - \ell \rvert$ $=\lvert 1+\frac{(-1)^n}{6 n}-1 \rvert = \frac{1}{6 n}$, and I want this to be $<\varepsilon$.

First guess: $k = \frac{1}{6\varepsilon}$.
 This gives $\frac1{6 k}\leqslant\varepsilon$ as required, but probably $\frac{1}{6\varepsilon} \notin \mathbb N$.
 So pick $k = \lceil\frac{1}{6\varepsilon}\rceil$.
 #### Summary
 If $n>\lceil\frac{1}{6\varepsilon}\rceil$, then I can show that $\lvert c_n - \ell \rvert<\varepsilon$.
 So I pick $k = \lceil\frac{1}{6\varepsilon}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: your turn 1
 Theorem:
 If $c_n = 1+\frac{(-1)^n}{6 n}$, then $\displaystyle\lim\_{n\to\infty} c_n = 1$.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert c_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert c_n - \ell \rvert$ $=\lvert 1+\frac{(-1)^n}{6 n}-1 \rvert = \frac{1}{6 n}$, and I want this to be $<\varepsilon$.

$\lvert c_n - \ell \rvert$ $= \lvert 1+\frac{(-1)^n}{6 n}-1 \rvert = \frac{1}{6 n}$ $<$ $\frac1{6 k} = \frac1{6\left\lceil1/6\varepsilon\right\rceil} \leqslant \frac1{6(1/6\varepsilon)} =$ $\varepsilon$.   □
 
 ---
 ## Deriving proof of sequence limit: your turn 2
 Theorem:
 If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.

Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert d_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert d_n - \ell \rvert$ $=?$

So  pick $k=?$
 
 ---
 ## Deriving proof of sequence limit: your turn 2
 Theorem:
 If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert d_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert d_n - \ell \rvert$ $=\lvert \frac{3+(-1)^n}{\sqrt n} - 0 \rvert = \frac{\lvert 3+(-1)^n \rvert}{\sqrt n} \leqslant \frac4{\sqrt n}$, and I want this to be $<\varepsilon$.

I may require $n>k$, so that $\frac1n<\frac1k$, so $\lvert d_n - \ell \rvert<$ $\frac4{\sqrt k}$, so now I just need $\frac4{\sqrt k}\leqslant\varepsilon$.

First guess: $k = \frac{16}{\varepsilon^2}$.
 This gives $\frac4{\sqrt k}\leqslant\varepsilon$ as required, but probably $\frac{16}{\varepsilon^2} \notin \mathbb N$.
 So pick $k = \lceil\frac{16}{\varepsilon^2}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: your turn 2
 Theorem:
 If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert d_n-\ell \rvert < \varepsilon$.

First guess: $k = \frac{16}{\varepsilon^2}$.
 This gives $\frac4{\sqrt k}\leqslant\varepsilon$ as required, but probably $\frac{16}{\varepsilon^2} \notin \mathbb N$.
 So pick $k = \lceil\frac{16}{\varepsilon^2}\rceil$.
 #### Summary
 If $n>\lceil\frac{16}{\varepsilon^2}\rceil$, then I can show that $\lvert d_n - \ell \rvert<\varepsilon$.
 So I pick $k = \lceil\frac{16}{\varepsilon^2}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: your turn 2
 Theorem:
 If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert d_n-\ell \rvert < \varepsilon$.

$\lvert d_n - \ell \rvert$ $= \lvert \frac{3+(-1)^n}{\sqrt n} - 0 \rvert = \frac{\lvert 3+(-1)^n \rvert}{\sqrt n} \leqslant \frac4{\sqrt n}$ $<$ $\frac4{\sqrt k} = \frac4{\sqrt{\left\lceil\frac{16}{\varepsilon^2}\right\rceil}} \leqslant \frac4{\sqrt{\frac{16}{\varepsilon^2}}} = \frac4{4/\varepsilon} =$ $\varepsilon$.   □