Thought process behind proving a limit of sequence

Question

My usual way of tackling "prove that $\alpha$ is the limit to the sequence $f(n)$ where $n\in \mathbb N$" is as follows.

First, I establish the necessary condition.

$|f(n)-\alpha|<\epsilon$ where $\epsilon>0$

Now, I work my way through to get the condition $n>g(\epsilon)$ and set $N= ceiling( g(\epsilon))$. Now I can state the proof in the traditional manner and it's done.

However, this approach doesn't seem to work for some problems. For instance, if I want to prove that $1/2$ is the limit of the sequence $\sqrt{n^2+n}-n$,

$|\sqrt{n^2+n}-n-1/2| \leq \sqrt{n^2+n}+n+1/2 < \sqrt{n^2+n}+n+1= \sqrt{n+1}(\sqrt{n}+\sqrt{n+1})<(n+1)(n+n+1)=(n+1)(2n+1)<\epsilon$

Now, according to the strategy I use I must find $g(\epsilon)$ such that $n>g(\epsilon)$ from the above result. But, I can't seem to do it.

My questions are,

Does the method I use always work?
If so, How can I modify my example to fit that strategy?

@YvesDaoust I just used $f(n)$ to show that its some function of $n\in \mathbb N$ — slhulk, May 11 '20 at 08:00
https://math.stackexchange.com/questions/783536/prove-that-lim-sqrtn2n-n-frac12 — A. Goodier, May 11 '20 at 08:02
@A.Goodier Thank you for directing me to the exact example in my question. I'm curious to know whether the method I use works every time. — slhulk, May 11 '20 at 08:04
@YvesDaoust oh! you mean when I used it for the second time. You're right. I corrected it. — slhulk, May 11 '20 at 08:20
Your goal is to ensure $|f(n) - L|<\epsilon$ and the technique is to find a simple $g(n) $ such that ensuring 1) $ |f(n) - L|<g(n) $ and 2) $g(n) <\epsilon $ are easy. You have found wrong $g(n) =(n+1)(2n+1)$ which makes 1) very easy and 2) impossible to ensure. Try finding a suitable $g(n) $. — Paramanand Singh, May 11 '20 at 14:10
Why not write $$n+1/2-\sqrt{n^2+n}=\frac{1}{4n+2+4\sqrt{n^2+n}}<1/4n$$ — Paramanand Singh, May 11 '20 at 14:15
@ParamanandSingh yes. Building a better $g(n)$ seems to work — slhulk, May 12 '20 at 02:16

score 1 · Accepted Answer · answered May 11 '20 at 08:34

You have to solve the equation

$$\forall n>N:|f(n)-\alpha|<\epsilon$$

for $N$. This is not exactly the same as finding $n$ such that

$$|f(n)-\alpha|<\epsilon$$ because $f$ might "oscillate" around $\alpha$.

In other terms, find the largest solution of

$$f(x)\in(\alpha-\epsilon,\alpha+\epsilon)$$ and take the ceiling.

If $f$ is monotonous around $\alpha$, this amounts to

$$x=\max(f^{-1}(\alpha-\epsilon),f^{-1}(\alpha+\epsilon)),$$ which is your $g(\epsilon)$. But if $f$ has a global minimum/maximum, the correct expression is

$$x=\max\left(\arg\min_{t\in(\alpha-\epsilon,\alpha+\epsilon)} f(t),\arg\max_{t\in(\alpha-\epsilon,\alpha+\epsilon)} f(t)\right).$$

In practice this approach can be complicated by the need to invert the function or discuss its extrema, and we often content ourselves with simpler, less tight bounds.

Thought process behind proving a limit of sequence

1 Answers1