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