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$\lim\limits_{n\to\infty }\sqrt {n^2+(-1)^n}-n$

First the domain , $n \geq 1$ since $n^2+(-1)^n \geq n^2-1 \geq 0$

let $a_n =\sqrt {n^2+(-1)^n}-n $ , so I wanted to simplify it by multiplying by the conjugate so $\sqrt {n^2+(-1)^n}+n \geq 0$ and I got

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

here is where I was not sure of how to use the squeeze theorem and this is what I did

$\frac{-1}{\sqrt{n^2-1}+n}\leq \frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n} \leq \frac{1}{\sqrt{n^2+1}+n} $

let $b_n = \frac{-1}{\sqrt{n^2-1}+n}$ and $c_n = \frac{1}{\sqrt{n^2+1}+n}$

so $\lim\limits_{n\to\infty }b_n = 0$ and $\lim\limits_{n\to\infty }c_n = 0$ and according to squeeze theorem $\lim\limits_{n\to\infty } a_n =0$

but in the book it was solved this way :

after multiplying by the conjugate they also got $\frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n}$ then they did $0 \leq|\frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n}| = \frac{|(-1)^n|}{|\sqrt{n^2+(-1)^n}+n|} = \frac{1}{\sqrt{n^2+(-1)^n}+n} \leq \frac{1}{n}$ and then they evaluated the limit $\lim\limits_{n\to\infty }0=0$ and $\lim\limits_{n\to\infty } \frac{1}{n} = 0$ so according to squeeze theorem $\lim\limits_{n\to\infty } a_n =0$

What I did not understand in the book is why out of no where they used an absolute value? when am I allowed to do that?

and is my try also correct? or do I need to explain more stuff because the inequality is not necessarily correct because for example in $C_n$ the denominator actually increases so the value decreases (although I think that in this case it doesn't matter because the value is positive while $b_n$ is negative ) but I mean in general ?

Thank you

EDIT:

I tried solving it in another way , as a multiplication of a limit equal to zero and a bounded limit. If I can separate this limit into 2 limits $\frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n}$ let $b_n=(-1)^n$ and $c_n =\frac{1}{\sqrt{n^2+(-1)^n}+n} $

$b_n$ is bounded of course $|b_n|=1 \lt 2 $ and $c_n =\frac{1}{\sqrt{n^2+(-1)^n}+n} \lt \frac{1}{n}$ because as stated in the beginning the domain , $n \geq 1$ since $n^2+(-1)^n \geq n^2-1 \geq 0$ from here we get $\sqrt{n^2-1} +n \geq n$

therefore $\lim\limits_{n\to\infty }b_n \cdot c_n=0$

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    $\begingroup$ There is a typo at the start when you multiply and divide by the conjugate. The square root sign shouldn't have the $-n$ under it. $\endgroup$
    – Digitallis
    Nov 8, 2021 at 14:42

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To address your question about how and why one can apply absolute values, here's a useful theorem:

A sequence $(a_n)$ converges to zero if and only if the corresponding sequence of absolute values $(|a_n|)$ also converges to zero.

So if your expectation is that $\lim_{n \to \infty} a_n = 0$, you can prove that by applying this theorem together with the squeeze theorem, in the following manner:

  • find another sequence $c_n$ such that $0 \le |a_n| \le c_n$ and such that $\lim_{n \to \infty} c_n = 0$;
  • apply the squeeze theorem to conclude that $\lim_{n \to \infty} |a_n| = 0$;
  • apply the theorem above to conclude that $\lim_{n \to \infty} a_n = 0$

Here's an even more general form of the above theorem that is also useful:

A sequence $(a_n)$ converges to $L$ if and only if $(a_n-L)$ converges to zero, if and only if $|a_n-L|$ converges to zero.

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  • $\begingroup$ appreciate the detailed reply it is much more understandable . so if I expect the limit to be zero I can apply $ 0 \leq |a_n| \leq c_n$ and solve accordingly. so the thing I tried is not correct I suppose and I should try as you just explained? $\endgroup$
    – Adamrk
    Nov 8, 2021 at 14:55
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    $\begingroup$ Yes, but keep in mind the comment of @Digitallis. $\endgroup$
    – Lee Mosher
    Nov 8, 2021 at 15:16
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    $\begingroup$ It is worth mentioning that the first statement is really just the squeeze theorem in disguise. Namely, if $\vert a_n \vert \leq c_n$, then we can equivalently write $$-c_n \leq a_n \leq c_n$$ and then apply the squeeze theorem. $\endgroup$ Nov 8, 2021 at 15:27
  • $\begingroup$ @LeeMosher Can you please check the small edit I made and tell me if this way is correct? $\endgroup$
    – Adamrk
    Nov 8, 2021 at 16:54
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    $\begingroup$ That looks fine. $\endgroup$
    – Lee Mosher
    Nov 8, 2021 at 16:58

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