HW Check: Prove that the sequence {a_n} converges to 1/2

[ironspud]

Hi,

Just hoping someone could check my work and point out any errors, if any.

Homework Statement

Consider the sequence {$a_n$} defined by $a_n=\frac{n}{2n+\sqrt{n}}$. Prove that $\lim_{x\to\infty}a_n=\frac{1}{2}$. (Do NOT use any of the "limit rules" from Section 2.2.)

Homework Equations

A sequence $a_n$ is said to converge to a real number A iff for each $\epsilon>0$ there exists a positive integer $n^*$ such that $\lvert a_n-A\rvert<\epsilon$ for all $n\geq n^*$.

The Attempt at a Solution

Proof:

Let $\epsilon>0$.

Let $n^*\in\mathbb{N}$ such that $n^*>[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2$.

If $n\geq n^*$, then

$\lvert a_n-A\rvert$

$=\lvert\frac{n}{2n+\sqrt{n}}-\frac{1}{2}\rvert$

$=\lvert\frac{2n-2n-\sqrt{n}}{2(2n+\sqrt{n})}\rvert$

$=\lvert\frac{-\sqrt{n}}{2(2n+\sqrt{n})}\rvert$

$=\lvert\frac{-n}{\sqrt{n}[2(2n+\sqrt{n})]}\rvert$

$=\lvert\frac{-n}{\sqrt{n}(4n+2\sqrt{n})}\rvert$

$=\lvert\frac{-n}{4n\sqrt{n}+2n}\rvert$

$=\lvert\frac{-n}{2n(2\sqrt{n}+1)}\rvert$

$=\lvert\frac{-1}{2(2\sqrt{n}+1)}\rvert$

$=\frac{1}{2(2\sqrt{n}+1)}$

$\leq\frac{1}{2\sqrt{n}+1}$

$<\frac{1}{2\sqrt{[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2}+1}$

$=\frac{1}{2\lvert[\frac{1}{2}(\frac{1}{\epsilon}-1)]\rvert+1}$

$=\frac{1}{2[\frac{1}{2}(\frac{1}{\epsilon}-1)]+1}$

$=\frac{1}{(\frac{1}{\epsilon}-1)+1}$

$=\frac{1}{\frac{1}{\epsilon}}$

$=\epsilon$.

 

[dynamicsolo]

...

$\lvert a_n-A\rvert$

$=\lvert\frac{n}{2n+\sqrt{n}}-\frac{1}{2}\rvert$

$=\lvert\frac{2n-2n-\sqrt{n}}{2(2n+\sqrt{n})}\rvert$

$=\lvert\frac{-\sqrt{n}}{2(2n+\sqrt{n})}\rvert$

You could have gotten here faster by dividing through by $\sqrt{n}$...

$=\lvert\frac{-1}{2(2\sqrt{n}+1)}\rvert$ ...

It looks all right to me. I am presuming that you come up with the "mysterious substitution" $\sqrt{[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2}$ for $\sqrt{n}$ by knowing that you're supposed to end up at "epsilon"...

 

[ironspud]

Thanks dynamicsolo.

And, yeah, I get the substitution by knowing $\frac{1}{2\sqrt{n^*}+1}<\epsilon$. Solving for epsilon, $[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2<n^*$. And since $n\geq n^*>[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2$, then $\frac{1}{2\sqrt{n}+1}<\frac{1}{2\sqrt{[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2}+1}$.

 

[dynamicsolo]

That's OK -- everyone does that in these epsilon-delta proofs. Your choice should be acceptable because $\epsilon(n) = \frac{1}{1 + 2\sqrt{n}}$ is one-to-one, so there is no ambiguity in transforming between epsilon and n , and the limit is zero as n goes to infinity.

 

[ironspud]

Yeah, I figured there was no problem in the substitution, though, maybe I should have provided the reasoning behind it in the proof. Oh well.

Anyway, thanks again.