Problem and solutions in convergent sequences
I will be sharing some solutions to problems on convergent sequences which i solved earlier today.
This is going to be one of the few posts i have written concerning convergent sequences so far


(1) check introduction to convergent sequence here
(2) click here for monotone convergent sequences 
(3) click here to see how to identify if a monotone sequence is convergent
(4) solutions to convergent sequences using epsilon limit approach

(1) Prove that $\frac{n^2}{2n^2+1}=\frac{1}{2}$
       
Solution:

To determine the convergency of this sequence is pretty simple, all we need to do is apply the definition of convergency i.e A sequence $\{a_n\}^\infty_n=1$ of $\mathbb{R}$ is said to converge to a real number $a_n$ if and only if for all $\epsilon>0$ there exists $\mathbb{N}$ such that $n(\epsilon):=|a_n-L|<\epsilon$ for all $n\geq{n(\epsilon)}$ where $n(\epsilon)$ is a natural number.
Now, we let

 $a_n=\frac{n^2}{2n^2+1}$ and $L=\frac{1}{2}$ so that

$|a_n-L|<\epsilon$
$\Rightarrow|\frac{n^2}{2n^2+1}-\frac{1}{2}|<\epsilon$

$\Rightarrow|\frac{2n^2-(2n^2+1)}{2(2n^2+1)}|<\epsilon\Rightarrow|\frac{-1}{2(2n^2+1)}|<\epsilon$

Now take the absolute value

$\frac{1}{2(2n^2+1)}<\epsilon$, now this is where most people get confused as to what to do next, we simply need to assume a value for $\epsilon$, to do this we multiply $\frac{1}{2}$ by $2n^2$ in the denominator and replace it by the constant which is $1$ i.e

$\frac{1}{2(2n^2+1)}<\epsilon=\frac{1}{2(2n^2+1)}<\frac{1}{2(\frac{2n^2}{2}+n^2)}$ for $\frac{2n^2}{2}>1\Rightarrow
n^2>1\Rightarrow{n>\pm{1}}$, this is the first value of $n$ satisfying $n\geq{n(\epsilon)}$.

$\Rightarrow{\frac{1}{2(2n^2+1)}<\frac{1}{2n^2+n^2}}=\frac{1}{2(3n^2)}$.

$\Rightarrow{\frac{1}{2(2n^2+1)}<\frac{1}{6n^2}}$

Now we must acknowledge the fact that
$6n^2<\epsilon\Rightarrow{n^2<\frac{\epsilon}{6}}$
$=n<\sqrt{\frac{\epsilon}{6}}$

This is the second value of $n$ satisfying $n\geq{n(\epsilon)}$, Hence we obtained maximum values for $n$ satisfying $n\geq{n(\epsilon)}$.

$n(\epsilon):=\max{[\pm+1],[\sqrt{\frac{\epsilon}{6}+1}]}$.


(2) Prove that $\lim_{n\rightarrow{\infty}}\frac{4n^3+3n}{n^3-6}=4$

Solution: By the definition of convergent sequence, i.e for all $\epsilon>0$ there exists $\mathbb{N}$ such that $n(\epsilon):=|a_n-L|<\epsilon$ for all $n\geq{n(\epsilon)}$ where $n(\epsilon)$ is a natural number.

Let $a_n=\frac{4n^3+3n}{n^3-6}$ and $L=4$
so that $n(\epsilon):=|\frac{4n^3+3n}{n^3-6}-4|<\epsilon$

$\Rightarrow|\frac{4n^3+3n-4(n^3-6)}{n^3-6}|<\epsilon$

$|\frac{4n^3+3n-4n^3+24}{n^3-6}|<\epsilon$

$|\frac{3n+24}{n^3-6}|<\epsilon\Rightarrow{\frac{3n+24}{n^3-6}<\epsilon}$

Now, we need to assume a value for $\epsilon$, to do this,  multiply $\frac{1}{2}$ by $n^3$ and substitute it in place of the constant $6$

$\frac{3n+24}{n^3-6}<\frac{3n+24}{n^3-\frac{n^3}{2}}$ for $\frac{n^3}{2}>6\Rightarrow{n^3}>12\Rightarrow{n>(12)^\frac{1}{3}}$ this is the first value of $n$ satisfying $n\geq{n(\epsilon)}$.

$\Rightarrow{\frac{3n+24}{n^3-6}<\frac{2(3n+24)}{2n^3-n^3}}$

$\Rightarrow{\frac{3n+24}{n^3-6}<\frac{2(3n+24)}{n^3}}$

Now, this is a problem that get most people hooked up, i.e when the numerator does not contain like terms, so all we need to do here is add $n$ to $24$ in the R.H.S

$\frac{3n+24}{n^3-6}<\frac{2(3n+24n)}{n^3}\Rightarrow{\frac{2(3n+24n)}{n^3}}$

$\frac{3n+24}{n^3-6}<\frac{2(27)}{n^2}$
$\Rightarrow{\frac{3n+24}{n^3-6}}<\frac{54}{n^2}$

Now that the Right hand side is fully evaluated, we let $\frac{54}{n^2}<\epsilon\Rightarrow{n}<\sqrt{\frac{54}{\epsilon}}$.

And this is the second value of $n$ satisfying $n\geq{n(\epsilon)}$ and we thus conclude the proof

$n(\epsilon):=\max\{[(12)^\frac{1}{3}],[\sqrt{\frac{54}{\epsilon}+1}]\}$.

And that bring us to the end of the solutions.






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