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Determine whether the following sequences are increasing or decreasing or neither.

a) $$\left(\ n + \sin (n) \right)_{n=0}^\infty$$ b) $$\left(\ n + \sin (1.047n) \right)_{n=1}^\infty$$

For: a) Let $f(x) = x +\sin(x) \implies f'(x) = 1 + \cos(x)$

$$ 1 + \cos(x)\geq 0 $$

and we also know that $$ -1 \leq \cos(x) \leq 1 $$

But since n starts at 1, the answer will be greater or equal to zero, regarding what n we take, therefore the function is increasing but i have no ideea how to prove that that the sequence is aswell. I know that i have to show that $$ \ a(n) \leq a(n+1) $$ but i don't exactly know how. Could you help me ?

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  • $\begingroup$ for all rational n, sin n >-1 $\endgroup$
    – Doug M
    Mar 20, 2016 at 21:47
  • $\begingroup$ "therefore the function is increasing but i have no ideea how to prove that that the sequence" The sequence is simply $a_n = f(n); n \in \mathbb Z$ If the function is increasing, the sequence, which is simply indexed values of the function is as well. $\endgroup$
    – fleablood
    Mar 20, 2016 at 21:57

1 Answer 1

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By the Mean Value Theorem, $\sin(n+1) - \sin(n) = \cos(t)$ for some $t$ between $n $ and $n+1$.

Actually , for (b) you'll want something better.

$$\sin((n+1)c) - \sin(nc) = (\cos(c) -1) \sin(nc) + \sin(c) \cos(nc)$$ so by Cauchy-Schwarz $$|\sin((n+1)c) - \sin(nc)| \le \sqrt{(\cos(c)-1)^2 + \sin(c)^2} = \sqrt{2 - 2 \cos(c)}$$

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