# Show that $\,\,\, \lim_{n\to\infty} \sin\bigl(\pi\sqrt{n^2+1}\bigr)=0$

Can anyone help me to solve this problem?

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

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Show that $\sqrt{n^2+1}\to n$ as $n\to\infty$. Also, $\sin(\pi n)=0\forall n\in \Bbb Z$ – Sabyasachi Mar 30 '14 at 15:32
@Sabyasachi: What you've written makes no sense. As $n\to\infty$, $\sqrt{n^2+1}\to\infty$. Do you want to say the difference between $\sqrt{n^2+1}$ and $n$ goes to $0$? The quotient goes to $1$? – Ted Shifrin Mar 30 '14 at 15:52
@TedShifrin I meant the quotient goes to $1$. The difference goes to $0$ as well I see. – Sabyasachi Mar 30 '14 at 15:53
Your claim is wrong take for example $a_n=n^2$ and $b_n=n^2+n$, the quotient goes to $1$ but the difference goes to $\infty$. @Sabyasachi – user63181 Mar 30 '14 at 15:59
I think Sabyasashi meant that $\;\sqrt{n^2+1}\sim n\;$ for big $\;n\;$, i.e.: the asymptotic behaviour of that square root is the same as $\;n\;$ . – DonAntonio Mar 30 '14 at 16:02

We have that $$0<\delta_n=\sqrt{n^2+1}-n=\frac{\big(\sqrt{n^2+1}+n\big)\big(\sqrt{n^2+1}-n\big)}{\sqrt{n^2+1}+n}= \frac{1}{\sqrt{n^2+1}+n}<\frac{1}{2n}$$ So $$\sin\left(\pi\sqrt{n^2+1}\right)=\sin\left(\pi\sqrt{n^2+1}-\pi n+\pi n\right)=\sin(\pi\delta_n+\pi n)=(-1)^n\sin(\pi\delta_n),$$ and as $\lvert \sin x\rvert\le \lvert x\rvert$, then $$\big|\sin\left(\pi\sqrt{n^2+1}\right)\big|=\lvert\sin(\pi\delta_n)\rvert\le \lvert\pi \delta_n\rvert<\frac{\pi}{2n}\to 0.$$ Note. In fact, one can use the above to show that $$\lim_{n\to\infty} (-1)^n n\sin\big(\pi\sqrt{n^2+1}\big)=\frac{\pi}{2}.$$

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We have by the Taylor series:

$$\sqrt{n^2+1}=n\sqrt{1+\frac1{n^2}}=n\left(1+O\left(\frac1{n^2}\right)\right)=n+O\left(\frac1{n}\right)$$ hence

$$\sin\left(\pi\sqrt{n^2+1}\right)=\sin\left(n\pi+O\left(\frac1{n}\right)\right)=(-1)^n\sin\left(O\left(\frac1{n}\right)\right)\xrightarrow{n\to\infty}\ 0$$

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have you assumed $n$ to be integer ? – lab bhattacharjee Mar 30 '14 at 15:40
Yes I usually consider $n$ to be integer. @labbhattacharjee – user63181 Mar 30 '14 at 15:42
does $n\to\infty \implies$ that ? Not sure what have meant by 'usually' – lab bhattacharjee Mar 30 '14 at 15:45
@Sami: lab's point is that the limit doesn't exist if you allow $n$ to be real instead of integer. So his question is where you are using that $n$ is an integer. The answer to that is in the last equal sign. – Martin Argerami Mar 30 '14 at 15:49
Yes I understood the lab's comment and my answer is: It seems very common to use $n$ for the expression of a sequence and not a function even this is not mentioned by the OP especially the question's tag is sequence and series .@MartinArgerami – user63181 Mar 30 '14 at 15:56

Assuming $n$ to be integer,

$$\sin\left(\pi\sqrt{n^2+1}\right)=(-1)^{n-1}\sin\left(n\pi-\pi\sqrt{n^2+1}\right)$$

$$=(-1)^{n-1}\sin\left[\pi(n-\sqrt{n^2+1})\right]$$

$$=(-1)^{n-1}\sin\left[\pi\left(\frac{n^2-(n^2+1)}{n+\sqrt{n^2+1}}\right)\right]$$

$$=(-1)^{n-1}\sin\left[-\pi\left(\frac1{n+\sqrt{n^2+1}}\right)\right]$$

Now $\displaystyle\lim_{n\to\infty}\left(\frac1{n+\sqrt{n^2+1}}\right)=0$

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$$\lim_{n\to\infty}\sin\pi\sqrt{n^2+1}=\lim_{n\to\infty}(-1)^n\sin(\sqrt{n^2+1}-n)\pi$$ $$=\lim_{n\to\infty}(-1)^n\sin(\frac{1}{\sqrt{n^2+1}+n})\pi=\lim_{n\to\infty}(-1)^n\sin0\cdot\pi$$ $$=0$$

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