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This fact is rather obvious (when viewed informally) but OK...
How can we prove most easily and without
using too much theory that the series

$$\sum_{n=1}^{\infty} \sin(n)$$

is divergent?

I tried to prove it but to do so, I have to use quite a few properties of the $sin$ and $arcsin$ functions. So I am looking for something simpler, more elegant.

peter.petrov
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1 Answers1

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New (hopefully improved) edit. Following the suggestion and the (fruitful and constructive) criticism in the comments, I have tried to prove directly the statement on the supremum and the infimum of the sequence $\{a_n\}_{n\in\Bbb N}=\{\sin(n)\}_{n\in\Bbb N}$: I think I succeeded in finding an entirely elementary proof, similar in the spirit to the ones shown in this question, so I have added it as a completion to my former answer.

Since $$ \liminf_{n\to\infty}a_n=-1\neq\limsup_{n\to\infty}a_n=1\label{1}\tag{1} $$ we have that $$ \lim_{n\to\infty}a_n=\lim_{n\to\infty}\sin(n)\neq 0\label{2}\tag{2} $$ thus the series does not converge since the necessary condition for convergence is not fulfilled.

Proof of the relation \eqref{1}. In order to do this, let's prove directly the above relation by proving the right side equality $\limsup_{n\to\infty}a_n=1$: this will be sufficient for two reasons, namely

  1. the method for proving the left side equality is completely analogous and,
  2. the right side equality is sufficient to guarantee that $\lim_{n\to\infty}a_n\neq 0$

Let's start recalling that $$ \sin\left[(4m+1)\frac{\pi}{2}\right]=1\quad \forall m\in\Bbb N.\label{3}\tag{3} $$ By applying the prostapheresis formulas to the difference $1-\sin(n)$, we get the following equality: $$ 0<1-\sin(n)=2\sin\left[\frac{(4m+1)\frac{\pi}{2}-n}{2}\right]\cos\left[\frac{(4m+1)\frac{\pi}{2}+n}{2}\right]\label{4}\tag{4} $$ Let's define an admissible range for $m$ as a function of $n$ by requiring the sine factor on the right side of \eqref{4} to be non negative: $$ \begin{align} 0&\le\sin\left[\frac{(4m+1)\frac{\pi}{2}-n}{2}\right]\le 1\\ &\Updownarrow \\ 0&\le \frac{(4m+1)\frac{\pi}{2}-n}{2} \le \pi\\ & \Updownarrow \\ [4(m-1)+1]\frac{\pi}{2} &\le n \le (4m+1)\frac{\pi}{2}\label{5}\tag{5} \end{align} $$ Since \eqref{5} implies the positivity of the sine factor, it also implies the positivity of \eqref{4}, therefore we get $$ 0<1-\sin(n)<2\sin\left[\frac{(4m+1)\frac{\pi}{2}-n}{2}\right] $$ For each $\epsilon> 0$ (and also sufficiently close to it, meaning that it must obviously be $1\ge \epsilon$), we now can define an infinite sequence $\{n_k\}_{k\in\Bbb N}$ such that $1-\sin(n_k)<\epsilon$: first note that it must be $$ \begin{split} 0<1-\sin(n_k)&<\epsilon\\ &\Updownarrow\\ 0<2\sin\left[\frac{(4m+1)\frac{\pi}{2}-n_k}{2}\right]&<\epsilon\\ &\Updownarrow\\ 0<{(4m+1)\frac{\pi}{2}-n_k}&<2\arcsin\Big(\frac{\epsilon}{2}\Big) \end{split} $$ i.e., due to \eqref{5}, $$ 0<{(4m+1)\frac{\pi}{2}-2\arcsin\Big(\frac{\epsilon}{2}\Big)} < n_k \le (4m+1)\frac{\pi}{2}. $$ Finally, by putting $$ m_\epsilon=\min\left\{m\in\Bbb N : (4m+1)\frac{\pi}{2}-2\arcsin\Big(\frac{\epsilon}{2}\Big)>0\right\} $$ we have the sought for sequence $$ n_k=\left\lfloor {(4(m_\epsilon+k)+1)\frac{\pi}{2}-2\arcsin\Big(\frac{\epsilon}{2}\Big)} \right\rfloor,\quad \forall k\in\Bbb N $$ and this implies $\limsup_{n\to\infty}\sin(n)=1$ by standard real analysis.

Daniele Tampieri
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    How do we know what the lim inf and lim sup are? $n$ takes only integer values. – saulspatz Sep 19 '19 at 13:15
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    @saulspatz They are the greatest and lowest upper bounds. For convergence it is necessary for the greatest upper and lowest upper bounds to be 0 as n tends to infinity. Furthermore the bounds come from the range of sin(n) (-1 beeing the smallest value it can take and 1 the biggest) – Maths Wizzard Sep 19 '19 at 13:20
  • @KanyeWest sup and inf are not the same as lim sup and lim inf. – saulspatz Sep 19 '19 at 13:21
  • Limsup = 1 (greatest upper value ) and Lim inf = (lowest upper value). The series is such that the sup = lim sup and inf = lim inf . This is why the series doesn't converge. – Maths Wizzard Sep 19 '19 at 13:23
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    @KanyeWest These statement require proof. – saulspatz Sep 19 '19 at 13:25
  • The proof is that we have better rational approximation of $ \pi /2$ by $m/n$, as $n$ tends to infinity. Hence $\sin n$ can be close to 1 and -1. –  Sep 19 '19 at 13:27
  • @LKM Yes, I know what the proof is. I' m not suggesting that the statements are false, but the OP has requested a proof. – saulspatz Sep 19 '19 at 13:37
  • This does not look at all simple to prove $$\liminf_{n\to\infty}a_n=-1\neq\limsup_{n\to\infty}a_n=1$$ I liked much more the idea to prove that sin(n) itself is not a convergent sequence. And that can be proved pretty easily as it seems. Thanks anyway. – peter.petrov Sep 19 '19 at 22:40
  • @peter.petrov, KayeWest, saulspatz, L KM. Thank you for your observations: evidently I missed the spirit of the question (as it appears also by having a look at the "duplicate" question). As L KM stated in his comment, the statement $\limsup a_n=1$ could be proved by showing that there exists a rational approximation $m/n$ of $\pi/2$ and $n\to \infty$ for $m$ properly chosen. I'll try to see if this can be done by elementary methods: if I will not succeed, tomorrow I'll delete my answer. – Daniele Tampieri Sep 20 '19 at 06:00