I would like to calculate the following limit: ${\lim _{n \to \infty }}\sin \left( {2\pi \sqrt {{n^2} + n} } \right)$
I am not sure if this limit exists...
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Is $n$ allowed to be any real, or are you restricting it to the integers? I think the limit doesn't exist for the former, and equals $0$ for the latter. – Akiva Weinberger Jul 24 '15 at 06:02
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1Helpful graph (The blue curve is the graph, and the red dots are the values at the whole numbers. The blue curve doesn't seem to approach a limit, but the red dots visibly go to zero.) – Akiva Weinberger Jul 24 '15 at 06:05
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4We have $\sqrt{n^2+n}=n+\frac{1}{2}+o(1)$. – André Nicolas Jul 24 '15 at 06:07
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Very closely related. – Jyrki Lahtonen Jul 24 '15 at 06:25
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Assuming that $n$ is an integer variable, $$\eqalign{ \sin(2\pi\sqrt{n^2+n}) &=\sin\bigl(2\pi\sqrt{n^2+n}-2\pi n\bigl)\cr &=\sin\left(\frac{2\pi}{\sqrt{1+\frac1n}+1}\right)\cr &\to\sin\pi\cr &=0\ .\cr}$$ If on the other hand $n$ is a real variable, then as $n$ tends to $\infty$, the expression $\sqrt{n^2+n}$ takes all positive real values, so $\sin(2\pi\sqrt{n^2+n})$ keeps on oscillating between $1$ and $-1$, and has no limit.
Jyrki Lahtonen
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David
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Thanks David for adding a line about the nature of $n$ and explaining the difference between real $n$ and integer $n$. People are so used to L'Hospital's Rule in limits that they differentiate to calculate limits involving integer $n$ also. +1 – Paramanand Singh Jul 24 '15 at 07:58
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Hint: Note that $$\sqrt{n^2+n}-\left(n+\frac{1}{2}\right)=\frac{\left(\sqrt{n^2+n}-\left(n+\frac{1}{2}\right)\right)\left(\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)\right)}{\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)}=\frac{-1/4}{\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)}.$$ Thus $$\sin\left(2\pi\sqrt{n^2+n}\right)=\sin\left(2\pi n+\pi - \frac{\pi/2}{\sqrt{n^2+n}+\left(n+\frac{1}{2} \right)}\right).$$
André Nicolas
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Hint: $2\pi \sqrt {n^2 +n} = 2\pi n\sqrt {1 +1/n}.$
zhw.
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It's a good start, but generally a one-line "hint" is not an Answer, unless the OP specifies that's what is wanted. Please expand your "hint" to show that it leads to a solution, or consider contributing this instead as a Comment. – hardmath Jul 24 '15 at 06:27
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You have a lot of work ahead of you to weed out one-line hints on these boards. Go to it. – zhw. Jul 24 '15 at 06:37
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With a limited development : $$\eqalign{ \sin(2\pi\sqrt{n^2+n}) &=\sin(2\pi n\sqrt{1+\frac1n})\cr &=_{n\rightarrow\infty}\sin(2\pi n\times(1+\frac1{2n}+o(\frac1n)))\cr &=\sin(2\pi \times(n+\frac12+o(1)))\cr &=\sin(\pi + 2\pi n + o(1))\cr &=-\sin(2\pi n + o(1)) \longrightarrow 0\ \cr}$$
BusyAnt
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Notice, let $n$ be any integer then we have $$\lim_{n\to \infty}\sin(2\pi\sqrt{n^2+n})$$ $$=\lim_{n\to \infty}\sin(-2n\pi+2\pi\sqrt{n^2+n})$$ $$=\lim_{n\to \infty}\sin\left(2\pi n\sqrt{1+\frac{1}{n}}-2\pi n\right)$$ $$=\lim_{n\to \infty}\sin\left(2\pi n \left(1+\frac{1}{n}\right)^{1/2}-2\pi n\right)$$ Since, $n$ is large enough hence using the binomial expansion of $\left(1+\frac{1}{n}\right)^{1/2}$ $$=\lim_{n\to \infty}\sin\left(2\pi n \left(1+\frac{1}{2}\left(\frac{1}{n}\right)+O\left(\frac{1}{n^2}\right)\right)-2\pi n\right)$$ $$=\lim_{n\to \infty}\sin\left(2\pi n \left(\frac{1}{2}\left(\frac{1}{n}\right)+O\left(\frac{1}{n^2}\right)\right)\right)$$ $$=\lim_{n\to \infty}\sin\left(\pi+(2\pi) O\left(\frac{1}{n}\right)\right)=\sin \pi=0$$ But, if $n$ is a real number then $\sin(2\pi\sqrt{n^2+n})$ can have any value within domain $[-1, 1]$ hence it has no definite limit (i.e. unique value)
Harish Chandra Rajpoot
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