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.