# Determine whether a sequence converge

• Aug 27th 2010, 05:57 PM
elim
Determine whether a sequence converge
Does $\displaystyle \{ |\sin n|^{\frac{1}{n}}\}$ converge?
• Aug 27th 2010, 07:13 PM
Prove It
Remember that

$\displaystyle -1 \leq \sin{n} \leq 1$ for all $\displaystyle n$, that means

$\displaystyle |\sin{n}| \leq 1$ for all $\displaystyle n$.

I assume since this is a sequence that $\displaystyle n$ only takes on postitive integer values, so that would mean $\displaystyle \frac{1}{n}$ is positive and therefore exponentiating would keep the inequality intact.

So $\displaystyle |\sin{n}|^{\frac{1}{n}} < 1^{\frac{1}{n}}$

$\displaystyle |\sin{n}|^{\frac{1}{n}} < 1$.

So yes, the sequence converges.
• Aug 27th 2010, 07:23 PM
elim
$\displaystyle |\sin n|^{\frac{1}{m}}<1$ not necessarily implies that it converges
• Aug 27th 2010, 07:28 PM
Prove It
True, I misinterpreted the question, sorry. All I proved was that it was bounded.

According to Wolfram,

$\displaystyle \lim_{n \to \infty}(|\sin{n}|^{\frac{1}{n}}) = 1$

so yes, the sequence converges.
• Aug 27th 2010, 08:29 PM
elim
Thanks a lot Sir!
what is the url at Wolfram?
• Aug 28th 2010, 04:39 AM
chisigma
Is...

$\displaystyle \displaystyle |\sin n|^{\frac{1}{n}} = e^{\frac{\ln |\sin n|}{n}$ (1)

... and because ...

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0$ (2)

... is ...

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} |\sin n|^{\frac{1}{n}} = 1$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Aug 29th 2010, 09:37 PM
elim
$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0$
is not obvious. there are integer $\displaystyle n's$ that make $\displaystyle |\sin n|$ very close to $\displaystyle 0$ hence $\displaystyle -\ln |\sin n|$ is very big.
• Aug 30th 2010, 01:33 AM
qspeechc
Do we have $\displaystyle |\text{sin} n|\leq n \quad \forall n\in\mathbb{N}$ ?
• Aug 30th 2010, 02:55 AM
Ackbeet
Reply to Elim at Post # 5:

Prove It is most likely referring to WolframAlpha. It's basically a way to enter single Mathematica commands. At least, that's what I use it for. Quite handy.
• Aug 30th 2010, 05:33 AM
elim
Quote:

Originally Posted by qspeechc
Do we have $\displaystyle |\text{sin} n|\leq n \quad \forall n\in\mathbb{N}$ ?

YES
• Aug 30th 2010, 05:35 AM
elim
Quote:

Originally Posted by Ackbeet
Reply to Elim at Post # 5:

Prove It is most likely referring to WolframAlpha. It's basically a way to enter single Mathematica commands. At least, that's what I use it for. Quite handy.

Thanks a lot Ackbeet. Very nice to know the online tool.
• Aug 30th 2010, 10:05 AM
chisigma
Quote:

Originally Posted by elim
$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0$
is not obvious. there are integer $\displaystyle n's$ that make $\displaystyle |\sin n|$ very close to $\displaystyle 0$ hence $\displaystyle -\ln |\sin n|$ is very big.

The consideration of elim is quite correct. By definition say that...

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0$ (1)

... means that for any $\displaystyle \varepsilon >0$ it exists an integer $\displaystyle N$ so that $\displaystyle \forall n>N$ is...

$\displaystyle \displaystyle \frac{|\ln |\sin n||}{n}< \varepsilon$ (2)

Now if we measure angles in degree [for example] instead of in radians $\displaystyle \forall n|180$ is $\displaystyle \sin n =0$ and the sequence $\displaystyle \displaystyle a_{n} = |\sin n|^{\frac{1}{n}}$ has no limit. If the angles are in radians then $\displaystyle \sin n$ never vanishes and in my opinion the (1) is 'probably true', even if for $\displaystyle \varepsilon$ 'very small' the (2) is satisfied for 'enormous' values of $\displaystyle N$. In any case a rigorous prove has to be supplied and that is not a trivial job (Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Aug 30th 2010, 11:17 AM
qspeechc
Quote:

Originally Posted by elim
YES

I don't mean to be irritating, but if you can show $\displaystyle n^{1/n}\rightarrow 0$ then we are done...
• Aug 30th 2010, 12:23 PM
Defunkt
Quote:

Originally Posted by qspeechc
I don't mean to be irritating, but if you can show $\displaystyle n^{1/n}\rightarrow 0$ then we are done...

$\displaystyle \displaystyle n^{1/n}\rightarrow 1$ :P
• Aug 31st 2010, 08:27 AM
qspeechc
Quote:

Originally Posted by Defunkt
$\displaystyle \displaystyle n^{1/n}\rightarrow 1$ :P

Oops. lol, I must have been smoking something potent...