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In order to show that $\lim_{n\to\infty}(1+\sin n)^{1/n}=1$, we need to show that $\,1+\sin n\,$ cannot not become too small. It suffices for example to show that $$ 1+\sin n\ge \frac{c}{n^k}, $$ for some $c,k>0$. This could be a consequence of showing that there exist $d,m>0$, such that $$ \Big|\,\pi-\frac{p}{q}\,\Big|\ge \frac{d}{q^{m}} $$ for every rational $p/q$ (cf. irrationality measure).

It would be very interesting if we could produce a more elementary proof.

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    The irrationality measure of $\pi$ is finite, so it is true. I'm not optimistic with respect to elementary proofs, though. – 2017-01-18
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    A first step may be to exploit the fact that the irrationality measure of $\pi$ is bounded by twice the irrationality measure of $\pi^2$, and the latter can be proved to be finite through Beuker-like integrals related with Legendre polynomials and the PNT. – 2017-01-18
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    You have to clarify how the sequence is defined. Be $N\in\mathbb{N}$ . If e.g. $n:=1+\pi N$ with $N\to\infty$ then the limit is $1$ . But with $n:=\frac{3\pi}{2}+2\pi N$ we have always $0$ and therefore the limit is $0$, and with $n:=\frac{\pi}{2}N$ we have no single limit and therefore divergence. – 2017-01-18
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    @user90369 In expressions with $\lim_{n\to\infty} x(n)$, the convention is that $n\in \mathbb{N}$, unless explicitly stated otherwise. – 2017-01-18
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    It seems to me that the two propositions are equivalent - if one can bound the value of $\sin n$ away from -1 effectively enough to yield the limit, then one could bound $\pi$ away from rational fractions effectively. It might be that there are other approaches to the question in the title, but given how closely the two translate back and forth it seems unlikely. – 2017-01-18
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    @Daniel Fischer. There is a very non-trivial result that $\pi$ (and hence any non-zero rational multiple of $\pi $) is not a Liouville number, which will imply that there exists $c>0$ and $k\geq 2$ such there are only finitely many $n\in \mathbb N$ for which $1+\sin n< cn^{-k}.$ – 2017-05-12

1 Answers 1

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Consider $f(x)=(1+\sin x)^{\frac{1}{x}}$ . The binomial expansion is valid for $|\sin x|<1$.

Case I

$x\neq\frac{\pi}{2}+2\pi k$ and $x\neq\frac{3\pi}{2}+2\pi k$, where $k\in\mathbb{Z}$. Hence we may expand binomially

$f(x)=1+\frac{\sin x}{x}+\frac{\frac{1}{x}(\frac{1}{x}-1)}{2!}\sin^2x+\dots+\frac{\frac{1}{x}(\frac{1}{x}-1)\dots(\frac{1}{x}-k+1)}{k!}\sin^kx+\dots$

EDIT (after comments made by Andréas)

$f(x)=1+\sum_{k=-x}^\infty\frac{(\frac{1}{x})!}{(x+k)!(\frac{1}{x}-x-k)!}\sin^{x+k}x$

Consider the coefficient of $\sin^{x+k}x$.

$g(x)=\frac{(\frac{1}{x})!}{(x+k)!(\frac{1}{x}-x-k)!}=\frac{\frac{1}{x}(\frac{1}{x}-1)\dots(\frac{1}{x}-x-k+2)(\frac{1}{x}-x-k+1)}{(x+k)!}$

Now taking the limit as $x\to\infty$,

\begin{align*} \lim_{x\to\infty}g(x) & =\lim_{x\to\infty}\frac{1}{x}\lim_{x\to\infty}\frac{(-1)(-2)\dots(-x-k+2)(-x-k+1)}{(x+k)!}\\ & =\lim_{x\to\infty}\frac{(-1)^{x+k-1}}{x}\lim_{x\to\infty}\frac{(x+k-1)!}{(x+k)!}\\ & =\lim_{x\to\infty}\frac{(-1)^{x+k-1}}{x(x+k)}\\ & =0 \end{align*}

Every term in the sum tends to zero, leaving $\lim_{x\to\infty}f(x)=1$.

Case II

$x=\frac{\pi}{2}+2\pi k$

Now $f(x)=2^{\frac{1}{x}}$ and clearly $\lim_{x\to\infty}f(x)=1$ again.

Case III

$x=\frac{3\pi}{2}+2\pi k$

Now $f(x)=0^{\frac{1}{x}}$ and therefore $\lim_{x\to\infty}f(x)=0$.

We may conclude that

$\lim_{x\to\infty}f(x)= \begin{cases} 0& x=\frac{3\pi}{2}+2\pi k\\ 1&\text{otherwise} \end{cases}$

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    This makes no sense. The right-hand side cannot be a function of $x $. – 2017-07-23
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    In case 1, you haven't explained why the limit is $0$. As written, it seems you forgot that $x$ depends on $k$, and treated $k$ as if it were constant. – 2017-07-23
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    Case 2 is actually correct, but irrelevant, because the problem is about integer $n$. For that matter, case 3 is also irrelevant. So, the only case that matters for the problem is case 1, where the argument you gave is (at best)incomplete. – 2017-07-23
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    @AndrésE.Caicedo $k$ is a constant. The $k^{\text{th}}$ term of the infinite series tends to zero for the same reason that $\lim_{x\to\infty}\frac{\sin x}{x}=0$. – 2017-07-23
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    No, $k $ is not constant. Consider, for instance, the case when $x=k $. :-) – 2017-07-23
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    It looks like what you're doing in case I is to write each $f(n)$ as a series indexed by $k$, and then taking the limit of these series term by term. But that is not valid: In general, $$ \lim_{n\to\infty}\sum_{k=0}^\infty a_{n,k} \quad\ne\quad \sum_{k=0}^\infty \lim_{n\to\infty} a_{n,k} $$ even when all the limits and sums exist. Consider for example $$a_{n,k}=\begin{cases}1&\text{if }n=k \\ 0 &\text{otherwise} \end{cases} $$ – 2017-07-23
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    @AndrésE.Caicedo I understand your objection now. There reaches a term in the infinite series where $k=n$. – 2017-07-23
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    @AndrésE.Caicedo I have now considered the infinite series with the dummy index $k$ being a function of $x$. – 2017-07-23
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    No, that argument does not work. It is not enough that each term goes to $0$. – 2017-07-23
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    @AndrésE.Caicedo Why is that so? If $\lim_{x\to x_0}A=\lim_{x\to x_0}B=\lim_{x\to x_0}C=0$, then $\lim_{x\to x_0}(A+B+C)=\lim_{x\to x_0}A+\lim_{x\to x_0}B+\lim_{x\to x_0}C=0$ – 2017-07-23
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    You are doing the same mistake you began with. You are not adding 3 terms, or any fixed finite number of terms. – 2017-07-23