How Does One Prove the Limit of the Sequence \(\sqrt{n^2+n}-n\)?

[ice109]

Homework Statement

find the limit and prove it using \epsilon and N(\epsilon) defn of limits f sequences, of this sequence:

x_n = \sqrt{n^2+n}-n

i don't know why I'm having so much trouble with this but anyway:

The Attempt at a Solution

let's say i divined that the limit is 1/2
we need to find N(\epsilon) such that n>N(\epsilon) implies

\big| \frac{1}{2} - \sqrt{n^2+n}-n \big| < \epsilon

well above implies

\big| \frac{1}{2} - \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n} \sqrt{n^2+n}-n \big| < \epsilon

which implies

\big| \frac{1}{2} - \frac{n}{\sqrt{n^2+n}+n} \big| < \epsilon

which implies

\big| \frac{1}{2} - \frac{n}{n\sqrt{1+\frac{1}{n}}+1} \big| < \epsilon

which implies

\big| \frac{1}{2} - \frac{1}{\sqrt{1+\frac{1}{n}}+1} \big| < \epsilon

which implies

\big| \frac{\sqrt{1+\frac{1}{n}}-1}{\sqrt{1+\frac{1}{n}}+1} \big| < 2\epsilon

which implies

\big| \frac{\sqrt{1+\frac{1}{n}}-1}{\sqrt{1+\frac{1}{n}}+1} \big| < \epsilon'

which implies

\big| \frac{\sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{\sqrt{1+\frac{1}{n}}+1}\big| < \epsilon'

but

\big| \frac{\sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{\sqrt{1+\frac{1}{n}}+1}\big| < \big| 1 - \frac{1}{\sqrt{1+\frac{1}{n}}}\big|

and by triangle inequality ( i think this is a mistake )

\big| 1 - \frac{1}{\sqrt{1+\frac{1}{n}}+1} \big| < 1 + \big| \frac{1}{\sqrt{1+\frac{1}{n}}} \big|

back to the epsilon, above implies we need to find n such that blah blah:1 + \big| \frac{1}{\sqrt{1+\frac{1}{n}}} \big| < \epsilon'

which implies

1+ \frac{1}{n} > \frac{1}{(\epsilon - 1)^2}

which implies:

n > \frac{ (\epsilon - 1)^2 }{1- (\epsilon - 1)^2}

so is this valid? intuitively it doesn't look immediately wrong to me? the term on the right is positive and increasing as epsilon gets smaller...

 

[tiny-tim]

Hi ice109!

(nice LaTeX, btw, but you can make it even nicer by typing \left| and \right| for big |s )

Sorry, I lost interest less than half-way through.

Why not use the same technique at the start, by putting √(n² + n) = n√(1 + 1/n)?

(or maybe compare it with y_n = √(n² + n + 1/4) - n )

 

[ice109]

Hi ice109!

(nice LaTeX, btw, but you can make it even nicer by typing \left| and \right| for big |s )

Sorry, I lost interest less than half-way through.

Why not use the same technique at the start, by putting √(n² + n) = n√(1 + 1/n)?

(or maybe compare it with y_n = √(n² + n + 1/4) - n )

if you hadn't lost interest you would have noticed i did do the first thing.

and your hint shows that my expression is less than 1/2 but it doesn't help me show that my sequence converges to 1/2 but that's cause i don't know quite how to use it to help me show that.

 

[aostraff]

I've lost interest as well, but I think your:

<br /> \big| \frac{1}{2} - \sqrt{n^2+n}-n \big| &lt; \epsilon <br />

should be:

<br /> \big| \frac{1}{2} - \left( \sqrt{n^2+n}-n \right) \big| &lt; \epsilon <br />