## Real Analysis – Sequences III

 Theorem 18 Theorem 17 Let ${ E}$ be a set of real numbers and ${ s=\sup E}$. Then there exists a sequence, ${ u_n}$,with range in ${ E}$ such as ${ \lim u_n=s}$. One can also formulate an analogous theorem for ${ i=\inf E}$. Proof: Omitted. $\Box$
 Theorem 19 Given the sequences ${ u_n}$ and ${ v_n}$ it is: ${ \lim u_n = a \in \mathbb{R}\Rightarrow \lim |u_n|=|a|}$ ${ \lim u_n=a \in \mathbb{R}}$ and ${ \lim v_n=b \in \mathbb{R}}$, then ${ lim (u_n+v_n)=a+b}$ ${ \lim u_n=+\infty}$ and ${ v_n}$ bounded below, then ${ \lim (u_n+v_n)=+\infty}$ ${ \lim u_n=-\infty}$ and ${ v_n}$ bounded above, then ${ \lim (u_n+v_n)=-\infty}$ ${ \lim u_n=0}$ and ${ v_n}$ bounded, then ${ \lim (u_n v_n)=0}$ ${ \lim u_n=a \in \mathbb{R}}$ and ${ \lim v_n=b \in \mathbb{R}}$, then ${ \lim u_n v_n = ab}$ ${ \lim |u_n|=+\infty}$ and ${ \lim v_n=a \neq 0}$, then ${ \lim |u_n v_n|= +\infty}$ ${ \lim u_n=a \in \mathbb{R}\setminus\{0\} \Rightarrow lim \dfrac{1}{u_n}=\dfrac{1}{a}}$ ${ \lim |u_n|=+\infty \Rightarrow \lim \dfrac{1}{u_n}=0}$ ${ \lim u_n=0 \Rightarrow \lim \dfrac{1}{|u_n|}=+\infty}$Proof: Only point 2 of the previous theorem will be proven. Let ${ \delta > 0}$. It is: $\displaystyle \begin{array}{rcl} |(u_n+v_n)-(a+b)| &=& |(u_n - a)+(v_n - b)| \\ &\leq & |u_n - a|+ |v_n - b| \end{array}$ Let the previous relationship be denoted by relationship 1 . We also have: ${ u_n \rightarrow a\Leftrightarrow \exists k_1 \in \mathbb{N}: \quad n \geq k_1 \Rightarrow |u_n - a| \leq \dfrac{\delta}{2}}$ ${ v_n \rightarrow b \Leftrightarrow \exists k_2 \in \mathbb{N}: \quad n \geq k_2 \Rightarrow |v_n - b| \leq \dfrac{\delta}{2}}$ Let ${ k=\mathrm{max}\{k_1,k_2\}}$ so that both the previous propositions are satisfied. Thus ${ \exists k \in \mathbb{N}: \quad n \geq k \Rightarrow |u_n - a|+|v_n - b|<\delta}$. Getting back to relationship 1 one finds: ${ |u_n - a|+|v_n - b| < \delta}$ and consequently ${ n \geq k \Rightarrow |(u_n + v_n)-(a+b)| < \delta}$. This is equivalent to ${ \lim (u_n + v_n) =a+b}$ if ${ \lim u_n = a}$ and ${ \lim v_n = b}$ which is the intended result. $\Box$

This theorem tell us the algebraic properties of limits of sequences and none of them seems to be too surprising to justify a presentation of all of the proofs.

Property 2 was shown in order for us to gain some more experience with the ${ k-\delta}$ notions.

In case you are wondering why we used ${ \dfrac{\delta}{2}}$ in the limit conditions of both ${ u_n}$ and ${ v_n}$ instead of ${ \delta}$ you have to realise that what matters in the definition of limit is that the distance between the terms of the sequence and its limit has to be smaller and smaller.

If we denote this distance by ${ \delta}$, or ${ \dfrac{\delta}{2}}$, or ${ \dfrac{\delta}{4}}$, or even ${ \displaystyle\exp\left( \sqrt{\dfrac{\delta+\phi}{\sqrt{3}}}\right)}$ is just a matter of convenience.

 Theorem 20 (Convergence of a monotone sequence) Let ${ u_n}$ be a monotone sequence in ${ \overline{\mathbb{R}}}$. Then ${ u_n}$ is a convergent sequence in ${ \overline{\mathbb{R}}}$: ${ \lim u_n = \mathrm{sup}\{ u_n: \, n \geq p \}}$ for ${ u_n}$ increasing. ${ \lim u_n = \mathrm{inf}\{ u_n: \, n \geq p \}}$ for ${ u_n}$ decreasing. Proof: Only the increasing case will be considered since the decreasing one is proved in a similar way. Given an increasing sequence ${ u_n}$, ${ E=\{ u_n:\, \geq p \}}$ and ${ s=\sup E}$. Let is first suppose ${ s \in \mathbb{R}}$ (a bounded above sequence). Given ${ \delta > 0}$ it is possible to prove that there exists ${ x \in E}$ such as ${ s-\delta < x\leq s}$. By the definition of ${ E}$ we know that ${ x=u_k}$, for a certain ${ k}$. Hence ${ \exists k \in \mathbb{N}:\quad s-\delta < u_k \leq s}$. Since ${ u_n}$ is an increasing sequence ${ n \geq k \Rightarrow u_n > s-\delta}$. But since ${ u_n \in E}$ we also have ${ u_n \leq s}$. Thus ${ n\geq k \Rightarrow s-\delta < u_n\leq s \Rightarrow u_n \in \rbrack s-\delta, s \rbrack \Rightarrow |u_n - s| < \delta}$. By definition of limit we have ${ \lim u_n = s}$. Let us now suppose ${ s=+\infty}$. In this case it can also be proven that given ${ L > 0}$ there exists ${ x \in E: \quad x > L}$. Remembering once again that it is ${ x=u_k}$ we have ${ \exists k \in \mathbb{N}: \quad u_k > L}$. Since ${ u_k}$ is an increasing sequence ${ n\geq k \Rightarrow u_n \geq u_k > L}$ and this equivalent to ${ u_n \rightarrow +\infty}$. $\Box$

Proper care is needed with the reading of the enunciate of this theorem. This is just a sufficient condition for a given sequence to have a limit. In no way the converse of the previous theorem (every sequence that has a limit in ${ \overline{\mathbb{R}}}$ is a monotone sequence) is a true statement. One has only to think about ${ u_n=\dfrac{(-1)^n}{n}}$ which tends to ${ 0 }$ and we see that it is not a monotone sequence even though it has a limit.

 Corollary 21 Every bounded and monotone sequence is convergent in ${ \mathbb{R}}$. Proof: By hypothesis we know that ${ \exists a,b \in \mathbb{R}: \, a\leq u_n \leq b}$. By the previous theorem we know that ${ u_n }$ has a limit in ${ \overline{\mathbb{R}}}$. And by the Corollary 15 it is ${ a \leq \lim u_n \leq b}$. Hence ${ u_n \rightarrow c \in \mathbb{R}}$ where ${ c \in \lbrack a, b \rbrack}$. $\Box$

Now this corollary here is a slobber-knocker for practical applications. Of course that, if the given sequence converges, we can’t say a thing about to what value it converges to.

But just think about the the fact the we are now able to determine the nature of a sequence without calculating any limits at all. It is now a matter of what is easier to do and/or what we need to know.

In some cases we may need the value of the limit but in other cases just knowing the behavior of the sequence is enough. And of course it also matters if it is more practical to actually calculate the limit and see if it exists and is a real number or not.

For instance given ${ u_n = \left( 1+\dfrac{1}{n} \right)^n}$ what should be our strategy? Go for the limit or try to prove that we have a bounded and monotone sequence?

Let us do some graphical inspection:

From the graph we can see that ${ u_n}$ appears to bounded by ${ 3}$ and is increasing, thus monotone.

I use the expression appears because a graphical representation can only contain a finite number of terms and one can’t be sure that something strange doesn’t happen at the points we aren’t representing.

But I’ll try to prove those two propositions anyway and conclude that ${ u_n}$ is indeed a convergent sequence.

 Proposition 22 ${ u_n = \left(1+\dfrac{1}{n}\right)^n}$ is a convergent sequence. Proof: First we’ll prove that ${ u_n }$ is increasing. In order to that we’ll calculate ${ u_{n+1}/u_n}$. {\begin{aligned} \dfrac{u_{n+1}}{u_n}&= \dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{\left(1+\dfrac{1}{n}\right)^n} \\ &=\dfrac{\left (\dfrac{n+2}{n+1} \right )^{n+1}}{\left ( \dfrac{n+1}{n} \right )^n} \\ &= \dfrac{n+2}{n+1}\left (\dfrac{n+2}{n+1}/\dfrac{n+1}{n} \right )^n \\ &= \dfrac{n+2}{n+1}\left (\dfrac{(n+2)n}{(n+1)^2} \right )^n \\ &= \dfrac{n+2}{n+1}\left (\dfrac{n^2+2n}{n^2+2n+1} \right)^n \\ &= \dfrac{n+2}{n+1} \left( \dfrac{n^2+2n+1-1}{n^2+2n+1} \right)^n \\ &= \dfrac{n+2}{n+1}\left (1-\dfrac{1}{n^2+2n+1} \right)^n \\ &= \dfrac{n+2}{n+1}\left (1-\dfrac{1}{(n+1)^2} \right )^n \end{aligned}} For us to proceed here we have to remember Bernoulli’s inequality ${ (1+x)^r \geq 1+rx\quad \forall r \in \mathbb{Z}}$ and ${ \forall x: \quad x \geq -1}$. This inequality can be proven (and will be in a future date) using mathematical induction. Continuing. {\begin{aligned} \dfrac{n+2}{n+1}\left (1-\dfrac{1}{\left (n+1 \right )^2} \right )^n &\geq \dfrac{n+2}{n+1}\left (1-\dfrac{n}{\left (n+1 \right )^2} \right ) \\ &= \left ( 1+\dfrac{1}{n+1}\right )\left ( 1-\dfrac{n}{\left(n+1\right)^2}\right ) \\ &= 1-\dfrac{n}{\left (n+1 \right )^2}+\dfrac{1}{n+1}-\dfrac{n}{\left (n+1 \right )^3} \\ &= 1-\dfrac{-n(n+1)+(n+1)^2-n}{\left (n+1 \right )^3} \\ &= 1+\dfrac{-n^2-n+n^2+2n+1-n}{\left (n+1 \right )^3} \\ &= 1+\dfrac{1}{\left (n+1 \right )^3} \\ &\geq 1 \end{aligned}} In conclusion ${ \dfrac{u_{n+1}}{u_n} \geq 1 \Leftrightarrow u_{n+1} \geq u_n}$ and ${ u_n }$ is monotone. Now all we have to do is to prove that ${ u_n}$ is bounded and we’ll know that is convergent. ${ u_1=\left(1+\dfrac{1}{1}\right)^1=2}$. From what we have already proven we also know that ${ u_n}$ is increasing, so ${ u_n \geq 2}$. Now we have to prove that ${ u_n}$ also has an upper bound for it to be bounded. As was proven in here we can write ${ \left(1+\dfrac{1}{n}\right)^n=\displaystyle \sum _{k=0}^n\dbinom{n}{k}\dfrac{1}{n^n}}$. Writing out the terms we have: {\begin{aligned} \left( 1 + \dfrac{1}{n}\right)^n &= \displaystyle \sum _{k=0}^n\dbinom{n}{k}\dfrac{1}{n^n} \\ &= 1+\dbinom{n}{1}\dfrac{1}{n}+\dbinom{n}{2}\dfrac{1}{n^2}+\cdots +\dbinom{n}{n}\dfrac{1}{n^n} \\ &= 1+1+\dfrac{n(n-1)}{2!}\dfrac{1}{n^2}+\cdots + \dfrac{1}{n^n} \end{aligned}} We know that ${ \dfrac{n(n-1)\ldots (n-(k-1))}{2!}\dfrac{1}{n^k}<1}$ and that ${ \dfrac{1}{k!}<\dfrac{1}{2^{k-1}}}$. Using those two inequalities (in that respective order) we have: { \begin{aligned} 1+1+\dfrac{n(n-1)}{2!}\dfrac{1}{n^2}+\cdots + \dfrac{1}{n^n} &< 1+1+\dfrac{1}{2!}+\cdots+\dfrac{1}{n!} \\ &< 1+\dfrac{1}{2}+\cdots+\dfrac{1}{2^{n+1}} \end{aligned} } So what we have is: ${ \left( 1+\dfrac{1}{n} \right)^n<\displaystyle 1 + \sum_{k=0}^{n-1}\left( \dfrac{1}{2} \right)^n}$. Since ${ \displaystyle\sum_{k=0}^{n-1}r^n=\dfrac{1-r^n}{1-r}}$ it is {\begin{aligned} \left( 1+\dfrac{1}{n} \right)^n &< 1+\dfrac{1-1/2^n}{1-1/2} \\ &= 1+\dfrac{1-1/2^n}{1/2} \\ &= 1+2-\dfrac{1}{2^{n-1}} \\ &= 3-\dfrac{1}{2^{n-1}} \\ &\leq& 3 \end{aligned}} Tidying up what we have is ${ 2\leq \left( 1+\dfrac{1}{n} \right)^n \leq 3}$. Hence ${ u_n}$ is monotone and bounded. Which amounts to ${ u_n}$ being convergent. Furthermore from ${ u_n \leq 3}$ we know that ${ \lim u_n \leq 3}$. This isn’t really much of a help to what value the sequence tends to but at least it is some information. $\Box$