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.