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I tried to use $\sin^2(\pi x) = \sin^2 (\pi x - \pi)$

So, $$\lim_{n\to\infty}\sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1}) \\ =\lim_{n\to\infty}\sin^2 (\pi(\sqrt{n^{2014} + n^{2012} + 1} - 1)) \\ \lim_{n\to\infty}\sin^2 (\frac{\pi (n^{2014} + n^{2012})}{\sqrt{n^{2014} + n^{2012} + 1}})$$

And I got trouble on here, Thanks for any hint.

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    $\begingroup$ What makes you think this is a convergent sequence? $\endgroup$ – Pspl Feb 13 at 15:52
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    $\begingroup$ The effect of $+1$ on the square-root is $O(n^{-1007})$ so it has no effect on the limit. Perhaps the Taylor series of $(1+n^{-2})^{1/2}$ tells you whether the full square-root is near an integer. $\endgroup$ – Empy2 Feb 13 at 16:24
  • $\begingroup$ @Pspl Because I saw a problem like this before and that was converge. so I tried solve this but I got trouble. $\endgroup$ – bFur4list Feb 13 at 16:53
  • $\begingroup$ I think I need to find some Polynomial $f(n)$ which satisfies : $\lim_{n\to\infty} (\sqrt{n^{2014} + n^{2012} + 1} - f(n)) = 0$. $\endgroup$ – bFur4list Feb 13 at 16:54
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NOTE: this is not a complete answer but a small collection of some partial results.

First remark

$$ \begin{gathered} \pi \sqrt {n^2 + n + 1} = \pi n\left( {1 + \frac{{\text{1}}} {n} + \frac{1} {{n^2 }}} \right)^{1/2} = \hfill \\ \hfill \\ = \pi n\left( {1 + \frac{1} {{2n}} + o\left( {\frac{1} {n}} \right)} \right) = \pi n + \frac{\pi } {2} + o\left( 1 \right) \hfill \\ \end{gathered} $$ this means that $$ \begin{gathered} \sin ^2 \left( {\pi \sqrt {n^2 + n + 1} } \right) = \sin ^2 \left( {\pi n + \frac{\pi } {2} + o\left( 1 \right)} \right) = \hfill \\ \hfill \\ = \sin ^2 \left( {\frac{\pi } {2} + o\left( 1 \right)} \right) \hfill \\ \end{gathered} $$ and therefore $$ \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^2 + n + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\frac{\pi } {2} + o\left( 1 \right)} \right)} \right] = 1 $$ This prove that, in general, it is not enough to consider only the main term of the polynomial under the square root.

Second remark Let us consider a more general case

$$ \begin{gathered} \pi \sqrt {n^{2k} + n^k + 1} = \pi n^k \left( {1 + \frac{{\text{1}}} {{n^k }} + \frac{1} {{n^{2k} }}} \right)^{1/2} = \hfill \\ \hfill \\ = \pi n^k \left( {1 + \frac{1} {{2n^k }} + o\left( {\frac{1} {{n^k }}} \right)} \right) = \pi n^k + \frac{\pi } {2} + o\left( 1 \right) \hfill \\ \end{gathered} $$ and as before we have that $$ \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^{2k} + n^k + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\frac{\pi } {2} + o\left( 1 \right)} \right)} \right] = 1 $$ On the other side, if we consider $n^{2k}+n^h+1$ with $0 \leq h<k$ we have that $$ \begin{gathered} \pi \sqrt {n^{2k} + n^h + 1} = \pi n^k \left( {1 + \frac{{\text{1}}} {{n^{2k - h} }} + \frac{1} {{n^{2k} }}} \right)^{1/2} = \hfill \\ \hfill \\ = \pi n^k \left( {1 + \frac{{\text{1}}} {{2n^{2k - h} }} + o\left( {\frac{{\text{1}}} {{n^{2k - h} }}} \right)} \right) = \pi n^k + \frac{\pi } {{2n^{k - h} }} + o\left( {\frac{1} {{n^{k - h} }}} \right) \hfill \\ \end{gathered} $$ thus $$ \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^{2k} + n^k + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\frac{\pi } {{2n^{k - h} }} + o\left( {\frac{1} {{n^{k - h} }}} \right)} \right)} \right] = 0 $$ Third remark: My conjecture about the case $k<h<2k$ is motivated by numerical trials like that below which is relative to the case $n^{20}+n^{18}+1$ enter image description here I have some ideas but not yet a proof. Of course, if my conjecture is true the proposed limit does not exists.

Fourth remark: Let be $h=k+1$ with $k \geq 3$. Then

$$ \begin{gathered} \pi \sqrt {n^{2k} + n^{k + 1} + 1} = \pi n^k \left( {1 + \frac{{\text{1}}} {{n^{k - 1} }} + \frac{1} {{n^{2k} }}} \right)^{1/2} = \hfill \\ \hfill \\ = \pi n^k \left( {1 + \frac{{\text{1}}} {{2n^{k - 1} }} - \frac{1} {{8n^{2k - 2} }} + o\left( {\frac{1} {{n^{2k - 2} }}} \right)} \right) = \hfill \\ \hfill \\ = \pi n^k + \frac{{\pi n}} {2} - \frac{1} {{8n^{k - 2} }} + O\left( {\frac{1} {{n^{k - 2} }}} \right) \hfill \\ \hfill \\ \end{gathered} $$ Therefore $$ \begin{gathered} \sin ^2 \left( {\pi \sqrt {n^{2k} + n^{k + 1} + 1} } \right) = \sin ^2 \left( {\pi \sqrt {n^{2k} + n^{k + 1} + 1} } \right) \hfill \\ = \sin ^2 \left( {\pi n^k + \frac{{\pi n}} {2} - \frac{1} {{8n^{k - 2} }} + O\left( {\frac{1} {{n^{k - 2} }}} \right)} \right) = \hfill \\ \hfill \\ = \sin ^2 \left( {\frac{{\pi n}} {2} - \frac{1} {{8n^{k - 2} }} + O\left( {\frac{1} {{n^{k - 2} }}} \right)} \right) \hfill \\ \hfill \\ \end{gathered} $$ so that the limit does not exists.

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  • $\begingroup$ Great! I'm gonna explore this just for the fun of it! $\endgroup$ – Pspl Feb 14 at 21:27
  • $\begingroup$ As soon as possible I will try to use more terms in the development of the square root. Right from now I'm quite sure that another term suffices to prove that the limit does not exists in the case $n^2k+n^k+1$ but I need time to write the complete proof of this case. Later I'll try to examine the full cases. $\endgroup$ – Luca Goldoni Ph.D. Feb 15 at 5:44
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I think the following argument is correct:

Imagine the sequence $u_n=\sin (\pi \sqrt{n^2+1}).$

It's easy to understand that $\lim_{n\to \infty}u_n=0$.

You can see it on the next representation of $u_n$:

u representation

That's because, when $n \to \infty$, you have $n^2+1 \sim n^2$ and you may write:

$$\lim_{n\to \infty}\sin (\pi \sqrt{n^2+1})=\lim_{n\to \infty}\sin (\pi \sqrt{n^2})=\lim_{n\to \infty}\sin (\pi n)=\lim_{n\to \infty}0=0$$

A similar argument may prove you that $\lim_{n\to\infty} \sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1})=0$, since $\sqrt{n^{2014} + n^{2012} + 1} \sim n^{1007}$ (when $n \to \infty$) which is always a natural number.

If there is a flaw on my argument, please let me know and I will delete this answer.

EDIT:

Actually, you can evaluate this using just the limits properties:

$$\begin{align*} \lim_{n\to\infty} \sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1})&=\sin^2 \left( \lim_{n\to\infty}(\pi\sqrt{n^{2014} + n^{2012} + 1}) \right)\\ &=\sin^2 \left(\pi \lim_{n\to\infty}(\sqrt{n^{2014} + n^{2012} + 1}) \right)\\ &=\sin^2 \left(\pi \sqrt{\lim_{n\to\infty}(n^{2014} + n^{2012} + 1}) \right)\\ &=\sin^2 \left(\pi \sqrt{\lim_{n\to\infty}n^{2014} } \right)\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;(1)\\ &=\sin^2 \left(\pi \lim_{n\to\infty}\sqrt{n^{2014} } \right)\\ &=\sin^2 \left(\pi \lim_{n\to\infty}{n^{1007} } \right)\\ &=\lim_{n\to\infty}\left(\sin^2 \left(\pi {n^{1007} } \right)\right)\\ &=\lim_{n\to\infty}\left(0^2\right)=\lim_{n\to\infty}0=0\\ \end{align*}$$

I used approximation before to justifying a known theorem $(1)$:

If $f(x)=a_0x^n+a_1x^{n-1}+ ... +a_n$ where $n\in \mathbb{N}$ and $a_0\neq0$, then: $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}a_0x^n$$

which you use on the simplifications above.

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  • $\begingroup$ This is one of reason which make me this limit converge, But I cannot find ways to solve this without approximation. $\endgroup$ – bFur4list Feb 13 at 17:06
  • $\begingroup$ Your argument is correct. However, in this problem one needs to look at the fractional part of $n^{1006}\sqrt{n^2+1}$, while the fractional part of $\sqrt{n^2+1}$ tends to $0$, the fractional part of $n^{1006}\sqrt{n^2+1}$ may not do the same. $\endgroup$ – user748968 Feb 13 at 17:07
  • $\begingroup$ @bFur4list, please take a look at the edited part of my answer. $\endgroup$ – Pspl Feb 13 at 17:43
  • $\begingroup$ @Pspl No, now what you added in the edit is not correct. The 'justification' of (1) is wrong. The two limits in the highlighted part are not finite and you want to compose these limits with $\sin$, which doesn't have a (unique) limit at $\infty$. $\endgroup$ – user748968 Feb 13 at 18:38
  • $\begingroup$ @tora, maybe you're right. But, at least, show us that, if the limit exists, is for sure equal to $0$. $\endgroup$ – Pspl Feb 13 at 20:46

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