sequence divergence proof

[shelly89]

$\displaystyle a_{n} = \sqrt{n} + 2sin (n)$

let m >0

take $\displaystyle N= (2+m)^{2}$ and let n >N

then

$\displaystyle \geq \sqrt{n} -2$

> 2+m-2 = m

because n >N

$\displaystyle \sqrt{n} > \sqrt{N} = 2+m$

where does N come from ?

we want $\displaystyle a_{n} > m$

start with

$\displaystyle a_{n} = \sqrt{n} + 2sin (n)$

$\displaystyle > \sqrt{n} -2$ which exceeds m

provided that

$\displaystyle \sqrt{n} > m+2$


provided

$\displaystyle n > (m+2)^{2}$


if

$\displaystyle N = (m+2)^{2}$



then

$\displaystyle n > N$

implies

$\displaystyle n > (m+2)^{2}$

therefore

$\displaystyle a_{n} > m$



can someone please explain how this proof to me i just cant follow it,

I dont understand how we go from

$\displaystyle \sqrt{n} +2sin(n) > m$

to

$\displaystyle > \sqrt{n} - 2$

how did they get this bit?

 

[daon2]

Note: If $\displaystyle 0<a<b$ then $\displaystyle \sqrt{a}<\sqrt{b}$.

If $\displaystyle (2+m)^2=N<n$, then $\displaystyle 2+m = \sqrt{(2+m)^2} = \sqrt{N} < \sqrt{n}$ so $\displaystyle m < \sqrt{n}-2$

 

[shelly89]

Note: If $\displaystyle 0<a<b$ then $\displaystyle \sqrt{a}<\sqrt{b}$.

If $\displaystyle (2+m)^2=N<n$, then $\displaystyle 2+m = \sqrt{(2+m)^2} = \sqrt{N} < \sqrt{n}$ so $\displaystyle m < \sqrt{n}-2$

thank you, but I still dont understand what happened to sin (n) ?

you go from

$\displaystyle \sqrt{n} -2sin(n)$

to

$\displaystyle \sqrt{n} -2$


how?

 

[JeffM]

thank you, but I still dont understand what happened to sin (n) ?

you go from

$\displaystyle \sqrt{n} -2sin(n)$

to

$\displaystyle \sqrt{n} -2$


how?

What is the largest value possible for 2 * sin(n) regardless of the value of n?

 

[shelly89]

What is the largest value possible for 2 * sin(n) regardless of the value of n?

well sin is bounded between -1 and 1,

 

[JeffM]

well sin is bounded between -1 and 1,

So $\displaystyle - 2 \le 2 * sin(n) \le + 2.$

Now do you see where the 2 came from and where the sine function went to?

 

[shelly89]

So $\displaystyle - 2 \le 2 * sin(n) \le + 2.$

Now do you see where the 2 came from and where the sine function went to?

understand where 2 'came from' but not where the sin (n) function 'went to', could you explain that bit ?


thanks

 

[pka]

where the sin (n) function 'went to', could you explain that bit ?

It did not 'go' anywhere.
$\displaystyle \\-1\le\sin(n)\le 1\text{, start with a well known fact}.\\ 2\ge -2\sin(n)\ge -2\text{, multiply by }-2 \\\sqrt{n}+2\ge \sqrt{n}-2\sin(n)\ge \sqrt{n}-2 \text{, add the }\sqrt{n} \text{ to all three members.}$

Do you see how it works?

 

[shelly89]

It did not 'go' anywhere.
$\displaystyle \\-1\le\sin(n)\le 1\text{, start with a well known fact}.\\ 2\ge -2\sin(n)\ge -2\text{, multiply by }-2 \\\sqrt{n}+2\ge \sqrt{n}-2\sin(n)\ge \sqrt{n}-2 \text{, add the }\sqrt{n} \text{ to all three members.}$

Do you see how it works?

oh goodness, now i get it! I just get very confused with rules of inequalities, its so confusing. But thank you , it all makes sense now.