Limit of $\sin^n(n)$

Question

Lately I thought about these limits:

$$\lim_{n\to\infty}\sin^n(n),\quad\lim_{n\to\infty}\cos^n(n)$$

Intuitively, I would say that they do not exist, but I wasn't able to prove it rigorously. Any help would be appreciated.

EDIT. No, the answer suggested does not answer my question, for two reasons: first, it's about convergence of the series

$$\sum_{n=1}^\infty\sin^nn$$

and, to prove that is does not converge, it is sufficient to show that, if the limit exists, it does not equal $0$. Hence, the answer only proves that, but doesn't prove that the limit doesn't exist (which would be done by proving the existence of a subsequence converging to $0$.) Secondly, I don't understand the proof provided anyway, I don't know if it's because it's written in a not very clear way or if it's me. Either way, it definitely doesn't answer my question.

Answer 1

Your question is already answered here. The key fact is Dirichlet approximation theorem (look at the proof by pigeonhole principle there).

Let me do the $(\cos(n))^n$ case. The idea is that if we have many pairs $(p,q)$ such that $\dfrac{p}{q}$ is very close to $\pi$, then $p$ is very close to $q\pi$, so $\cos(p)$ is very close to $\cos(q\pi)$ which is $\pm 1$, and if it's really close to $1$ then $(\cos(p))^p$ would still be fairly close to $1$.

In detail, by Dirichlet approximation theoerm there are infinitely many pairs $(p,q)$ of coprime integers satisfying $|\dfrac{p}{q}-\pi|\leq\dfrac{1}{q^2}$, in other words $|p-q\pi|\leq\dfrac{1}{q}$. Say $q$ is even so $\cos(x+q\pi)=\cos(x)$. Using the inequality $\cos(x)\geq 1-\dfrac{x^2}{2}$ we have

$$\cos(p)=\cos(p-q\pi+q\pi)=\cos(p-q\pi)\geq 1-|p-q\pi|^2\geq 1-\dfrac{1}{q^2}$$

and therefore

$$(\cos(p))^p\geq\left(1-\frac{1}{q^2}\right)^p\geq\left(1-\frac{1}{q^2}\right)^{q\pi-\frac{1}{q}}$$

It is not hard to see that $\left(1-\frac{1}{q^2}\right)^{q\pi-\frac{1}{q}}$ tends to $1$. The case when $q$ is odd is similar ($\cos(p)$ would be close to $-1$). Since either the even case or the odd case has to happen infinitely many times, there is a subsequence of $(\cos(n))^n$ that tends to either $1$ or $-1$. Now clearly there also exists a subsequence that tends to zero, so the sequence diverges.

Answer 2

See, let's take the examples one by one. That is let's first assume the limit $$\lim_{x\to \infty}[\sin(x)]^{x}$$

Now, we all know that $$-1\leq \sin(x)\leq 1$$

Therefore $$-1\leq \lim_{x\to \infty}\sin(x)\leq 1$$

Now, here lies the twist. Please try to understand carefully.

We all know that $$\lim_{n\to \infty}x^{n}$$ is $=0$ when $0<x<1$

And $$\lim_{n\to \infty}x^{n}$$ is undefined when $-1<x<0$

Why undefined because you can't predict the sign between $+$ and $-$ at $n\to \infty$.

Therefore $$\lim_{x\to \infty}[\sin(x)]^{x}$$ is equal to $0$ for $0<\sin(x)<1$.

And $$\lim_{x\to \infty}[\sin(x)]^{x}$$ is equal to undefined for

$-1\leq\sin(x)<0$.

Because you can't predict the sign between $+$ and $-$

Because, we all know about the example $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}$$

Here for $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}$$ at $x=0$, the value is predictable for $(1+x)$. So, we can apply the calculations for $1^{\infty}$.

But for $$\lim_{x\to \infty}[\sin(x)]^{x}$$ the value is not predictable at $x=\infty$ because we can't say directly that $\sin(\infty)=1$

But if, $$\lim_{x\to \infty}[\sin(x)]^{x}$$ is of the form of $1^{\infty}$, when $\lim_{x\to \infty}\sin(x)=1$ then apply this rule :

If $$L=f(x)^{g(x)}$$ where $f(x)=1$ and $g(x)=\infty$ at any approaching value of $x$, then we can say that

$$L=e^{g(x)[f(x)-1]}×\lim_{x\to k}f(x)$$

Where $k$ is the Approaching value for which $f(x)=1$ and $g(x)=\infty$

Try for $[\cos(x)]^{x}$ by the same process.

Answer 3

If a sum does not converge, it does not imply that $\lim\limits_{n→∞} T_n$ is not zero.

The limit $\sin^n(n)$ as $n$ tends to infinity does not exist. To understand that we need to understand limits intuitively, particularly where the input tends to infinity. I think of it as "as we put higher and higher values of $n$, the function should tend to a certain constant value and the change in the value of output should be in degree(s) of $n$ which are less than zero."

Simply put if $\lim\limits_{n→∞} f(n)$ exists, then $\lim\limits_{n→∞} f '(n) = 0$.

In the given function, the derivative varies periodically so the limit does not exist.

Answer 4

I will start by focusing on $\sin^n(n)$ and leave the cosine case to you. Consider the subsequence where $n =\pi m$ where $m$ is an integer. Then we have that: $$ \sin^n(n) = (0)^{\pi m} =0 $$ Now consider the subsequence where $n = \pi/2 + 2\pi \ell$. In this case: $$ \sin^n(n) = (1)^{\pi/2 + 2\pi \ell} =1 $$ Now because these two subsequences converge to different values the limit does not exist.