Sequences and bounds

## Sequences and bounds
            ---
            ## Sequences: motivation
            $(x,y)$,   $(x,y,z)$ are ordered pair and ordered triplet.

$(a,b,c,d,e,f,g)$ is an ordered heptuplet.

$(a_1,a_2,a_3,\ldots,a_n)=(a_j)_{j=1}^n$ is an ordered $n$-tuplet.

$(a_1,a_2,a_3,\ldots)$, usually written $(a_j)\_{j=1}^\infty$ is an infinite *sequence*.
            ---
            ## Sequences: Examples
            * $(7,14,21,28,35,\ldots) = (7n)\_{n=1}^\infty$
            * $(a_n)\_{n=1}^\infty$ defined by $a\_n = \frac{(-1)^n}n$
            * $(c_n)\_{n=1}^\\infty$ where $c\_{2j}=j^2$ and $c\_{2j-1}=\frac{1}{j^2}$

Then

$(a_1, a_2, a_3, a_4, a_5, a\_{100}) = $ ?

$(c_1, c_2, c_3, c_4, c_5, c\_{100}) = $ ?
            
            ---
            ## Sequences: Examples
            * $(7,14,21,28,35,\ldots) = (7n)\_{n=1}^\infty$
            * $(a_n)\_{n=1}^\infty$ defined by $a\_n = \frac{(-1)^n}n$
            * $(c_n)\_{n=1}^\\infty$ where $c\_{2j}=j^2$ and $c\_{2j-1}=\frac{1}{j^2}$

Then

$(a_1, a_2, a_3, a_4, a_5, a\_{100}) = (-1,\frac12,\frac{-1}3,\frac14,\frac{-1}5,\frac1{100})$

$(c_1, c_2, c_3, c_4, c_5, c\_{100}) = (1,1,\frac14,4,\frac19,2500)$
            
            ---
            ## Bounded sequences
            #### Definition
            A sequence $(a_n)\_{n=1}^\infty$ is *bounded* if<br>there exists some $M\geq0$ such that, for all $n\in\mathbb N$, $\lvert a_n \rvert \leq M$.

We call $M$ a *bound* for $(a_n)\_{n=1}^\infty$.
            #### Examples
            If $a_n = \frac3n$, then $3$ is a bound for $(a_n)\_{n=1}^\infty$.

If $b_n = \frac{3n}{n^2+1}$, then $5$ is a bound for $(b_n)\_{n=1}^\infty$, and so is $2$.

If $c_n = -n^3$, then what is a bound for $(c_n)\_{n=1}^\infty$?

If $d_n = 1-\frac1n$, then what is a bound for $(d_n)\_{n=1}^\infty$?
            #### Smallest bound
            We do not need the smallest bound for this course; see MATH2350.
            
            ---
            ## Proof of a bound
            #### Theorem
            If $d_n = 1-\frac1n$, then $1$ is a bound for $(d_n)\_{n=1}^\infty$.
            #### Proof
            We aim to show that, for all $n\in\mathbb N$, both $-1 \leqslant 1-\frac1n$ and $1-\frac1n \leqslant 1$.
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:1;margin-right:2vw">

\\[\begin{aligned}
              & & n & \geqslant 1 \\\\
              & \implies & 1 & \geqslant \frac1n  \\\\
              & \implies & -1 & \leqslant -\frac1n  \\\\
              & \implies & 0 & \leqslant 1-\frac1n
            \end{aligned} \\]
            </div>
            <div style="flex:1;margin-left:2vw">

\\[\begin{aligned}
              & & 1 & > 0 \\\\
              & \implies & \frac1n & > 0  \\\\
              & \implies & -\frac1n & < 0 \\\\
              & \implies & 1-\frac1n & < 1
            \end{aligned} \\]
            </div>
            </div>

We have shown stronger statements than we needed.
            
            ---
            ## Deriving the proof of a bound
            We aim to show that, for all $n\in\mathbb N$,  $-1 \leqslant 1-\frac1n$.
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:1;margin-right:2vw;border-right: 2px solid var(--r-main-color);">

The final proof is
            \\[\begin{aligned}
              & & n & \geqslant 1 \\\\
              & \implies & 1 & \geqslant \frac1n  \\\\
              & \implies & -1 & \leqslant -\frac1n  \\\\
              & \implies & 0 & \leqslant 1-\frac1n
            \end{aligned} \\]
            </div>
            <div style="flex:1;margin-left:2vw;border-right: 2px solid var(--r-main-color);">

But to work this out we do
            \\[\begin{aligned}
              & & -1 & \leqslant 1-\frac1n \\\\
              & \implies & \frac1n & \leqslant 2  \\\\
              & \implies & \frac12 & \leqslant n,
            \end{aligned} \\]
            which is true.
            </div>
            <div style="flex:1;margin-left:2vw">

Then make this a proof by
            \\[\begin{aligned}
              & & -1 & \leqslant 1-\frac1n \\\\
              & \impliedby & \frac1n & \leqslant 2  \\\\
              & \impliedby & \frac12 & \leqslant n,
            \end{aligned} \\]
            which is true.
            </div>
            </div>

Then we can reorder and tighten the bound if we want, but it is not required.
            
            ---
            ## Deriving the proof of a bound: your turn
            We aim to show that, for all $n\in\mathbb N$,  $1-\frac1n \leqslant 1$.
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:1;margin-right:2vw;border-right: 2px solid var(--r-main-color);">

The final proof is
            \\[\begin{aligned}
              & & 1 & > 0 \\\\
              & \implies & \frac1n & > 0  \\\\
              & \implies & -\frac1n & < 0 \\\\
              & \implies & 1-\frac1n & < 1
            \end{aligned} \\]
            as required.
            </div>
            <div class="fragment daz-switch preload" data-fragment-index="0" style="flex:1;margin-left:2vw;border-right: 2px solid var(--r-main-color);">

But to work this out we do
            \\[\begin{aligned}
              & & 1-\frac1n & \leqslant 1 \\\\
              & \implies & &? \\\\
              & & &\phantom{\leqslant 0,}
            \end{aligned} \\]
            </div>
            <div class="fragment daz-switch" data-fragment-index="1" style="flex:1;margin-left:2vw;border-right: 2px solid var(--r-main-color);">

But to work this out we do
            \\[\begin{aligned}
              & & 1-\frac1n & \leqslant 1 \\\\
              & \implies & -\frac1n & \leqslant 0 \\\\
              & \implies & \frac1n & \geqslant 0 \\\\
              & \implies & 1 & \geqslant 0,
            \end{aligned} \\]
            which is true.
            </div>
            <div class="fragment daz-switch preload" data-fragment-index="0" style="flex:1;margin-left:2vw">

Then make this a proof by
            \\[\begin{aligned}
              & & 1-\frac1n & \leqslant 1 \\\\
              & \phantom{\implies} & & ? \\\\
              & & &\phantom{\leqslant 0,}
            \end{aligned} \\]
            </div>
            <div class="fragment daz-switch" data-fragment-index="1" style="flex:1;margin-left:2vw">

Then make this a proof by
            \\[\begin{aligned}
              & & 1-\frac1n & \leqslant 1 \\\\
              & \impliedby & -\frac1n & \leqslant 0 \\\\
              & \impliedby & \frac1n & \geqslant 0 \\\\
              & \impliedby & 1 & \geqslant 0,
            \end{aligned} \\]
            which is true.
            </div>
            </div>

Then we can reorder and tighten the bound.
            
            ---