Solving Convergent Sequences Using Epsilon or Limit Approach

Solution To Problems of Convergence Sequence using Epsilon Approach

In one of my previous post titled "The convergence Sequence" i discussed about how to identify sequences that are convergent and also gave an example of how to solve a sequence that approaches or converges to a particular limit using a very simple method of limit at infinity.
Below i will be solving some sequence problems using another method of convergent sequence known as the "epsilon or limit approach".

Example 1
prove that $\lim_{n\rightarrow\infty}\frac{n}{n+1}=1$

Solution: using epsilon or limits definition of convergency which says that for all $\epsilon>0$ there is a natural number $\mathbb{N}$ such that $|s_n-L|<\epsilon$ for all $n\geq{\mathbb{N}(\epsilon)}$.
Now let $s_n=\frac{n}{n+1}$ and $L=1$ then $$\begin{align*} &=|\frac{n}{n+1}-1|<\epsilon\\ &=|\frac{n-(n+1)}{n+1}|<\epsilon\\ &=|\frac{-1}{n+1}|<\epsilon\\ &=\frac{1}{n+1}<\epsilon \end{align*}$$
since $\epsilon>0$ and it is a very small positive unknown number we let $\epsilon=\frac{1}{n}$ so that
$\frac{1}{n+1}<\frac{1}{n}$ and we must still note that $\frac{1}{n}<\epsilon$ subsequently $n>\frac{1}{\epsilon}$ now since $n\geq{\mathbb{N}(\epsilon)}$ then we can finalise our answer as $$\begin{align*} \mathbb{N}(\epsilon)=[\frac{1}{\epsilon}+1] \end{align*}$$
therefore $\mathbb{N}(\epsilon):|s_n-L|<\epsilon$ for all $n\geq{\mathbb{N}(\epsilon)}:[\frac{1}{\epsilon}+1]$

Example 2
Prove that $\lim_{n\rightarrow{\infty}}\frac{1}{n^2}=0$

Solution: we are being asked to show that the given sequence approaches zero(0) as the limit n approaches $infinity(\infty)$.
using the epsilon or limit approach,let the sequence $s_n=\frac{1}{n^2}$ and the limit $L=0$ so that when we substitute inside the limit formula we get
$$\begin{align*} &|s_n-L|<\epsilon\\ &=|\frac{1}{n^2}-0|<\epsilon\\ &=|\frac{1}{n^2}|<\epsilon\\ &=\frac{1}{n^2}<\epsilon \end{align*}$$
since the epsilon is a very small positive unknown number we let
$\epsilon=\frac{1}{n}$ such that $\frac{1}{n^2}<\frac{1}{n}$ where $n=2$ now $\frac{1}{n}<\epsilon$ make n subject of formula such that $n>\frac{1}{\epsilon}$ now since we know by definition that $n\geq{\mathbb{N}(\epsilon)}$
therefore $\mathbb{N}(\epsilon):=[\frac{1}{\epsilon}+1]$ which is the convergency of the sequence.

Example 3
Show  that $\lim_{n\rightarrow\infty}\frac{3n+1}{7n-4}=\frac{3}{7}$
Solution:
  $\lim_{n\rightarrow\infty}\frac{3n+1}{7n-4}=\frac{3}{7}$ meaning the sequence $\frac{3n+1}{7n-4}$ converges to $\frac{3}{7}$ as the limit n approaches infinity.
By definition of convergency we know that every convergent sequence must satisfy $\mathbb{N}(\epsilon):|s_n-L|<\epsilon$ for every $\epsilon>0$ and $n\geq{\mathbb{N}(\epsilon)}$ Let $s_n=\frac{3n+1}{7n-4}$ and $L=\frac{3}{7}$
$$\begin{align*} |\frac{3n+1}{7n-4}-\frac{3}{7}|<\epsilon \end{align*}$$
take the L.C.M of the denominator which $7(7n-4)$,

$$\begin{align*} &|\frac{7(3n+1)-3(7n-4)}{7(7n-4)}|<\epsilon\\ &=|\frac{21n+7-21n+12}{7(7n-4)}|<\epsilon\\ &=|\frac{19}{7(7n-4)}|<\epsilon\\ &=\frac{19}{7(7n-4)}<\epsilon \end{align*}$$
multiply $\frac{1}{2}$ by what carries "n" in the denominator such it is greater than the constant number i.e $\frac{7n}{2}>4$.
$$\begin{align*} \frac{19}{7(7n-4)}<\frac{19}{7(7n-\frac{7n}{2})} \end{align*}$$
$\frac{7n}{2}>4$ if n=2 we get $7>4$
$$\begin{align*} &\frac{19}{7(7n-\frac{7n}{2})}\\ &=\frac{19}{7(\frac{14n-7n}{2})}\\ &=\frac{38}{49n}<\epsilon\\ &=n>\frac{38}{49\epsilon} \end{align*}$$
But remember that $\frac{7n}{2}>4=7n>8\Rightarrow{n}>\frac{8}{7}$
Since $n\geq{\mathbb{N}(\epsilon)}$ therefore
$$\begin{align*} n\geq{\mathbb{N}(\epsilon)}:=max\{(\frac{8}{7}+1)(\frac{38}{49\epsilon}+1)\} \end{align*}$$
the covergency is proved, the answer is maximum because two values of $n$ was obtained at the end of the solution.

Play Question
Show whether or not $\lim_{n\rightarrow\infty}\frac{(-1)^n}{n}=0$ is convergent using any method possible.