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I have the following sequence:

$$a_n = \frac{\sin(n)+\cos(n)}{\sqrt n}$$

I have to find a $N(\epsilon)$ that $a_n$ convergate for an arbitrary $\epsilon > 0$.

The definition is: $\forall \epsilon > 0 , \; \exists N(\epsilon) \in \mathbb{N} , \; \forall n > N(\epsilon):\vert a_n - a \vert$

I started with setting into the definition:

My assumption is that $0$ is my limit.

$$\left| \frac{\sin(n)+\cos(n)}{\sqrt n} - 0 \right| < \epsilon =\sin(n) +\cos(n) < \epsilon \sqrt n =?$$

Here I get stucked because I need to have all $n$ on one side.

Thank you!

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  • $\begingroup$ Hint: $\sin(x)$ and $\cos(x)$ are bounded, so you can get rid of the $n$ on one side. $\endgroup$ – Ingix Mar 8 '17 at 8:29
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Since $|\sin(n)| \le 1$ and $|\cos(n)| \le 1$ we get:

$|a_n-0| \le \frac{2}{\sqrt{n}}$.

$\frac{2}{\sqrt{n}} < \epsilon$ $ \iff$ $n> \frac{4}{\epsilon^2}$

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  • $\begingroup$ While $\sin(n)$ and $\cos(n)$ are both individually between -1 and 1, their sum might not be. The easiest way to bound that sum would be to use the triangle inequality: $|\sin(n)+\cos(n)| \le |\sin(n)|+|\cos(n)| \le 1+1 = 2$. This results in $|a_n-0| \le \frac{2}{\sqrt{n}}$ and finally $n>\frac{4}{\epsilon^2}$ $\endgroup$ – Ingix Mar 8 '17 at 8:42
  • $\begingroup$ Ooops you are right. An edit will follow $\endgroup$ – Fred Mar 8 '17 at 8:47
  • $\begingroup$ Thank you for you answers. I am new to this proof stuff so I don't really understand it correctly I think. So the stuff with the triangle inequality is what I understand, but how did you follow $ \vert sin(n) \vert + \vert cos(n) \vert \le 2 => \vert a_n - 0 \vert \le \frac{2}{\sqrt n} => n \ge \frac{4}{\epsilon^2}$ It is not obvious for me. $\endgroup$ – flow Mar 8 '17 at 15:17
  • $\begingroup$ Ok now I got it I think, thank you very much. $\endgroup$ – flow Mar 8 '17 at 15:30

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