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I'm trying to investigate the boundness of the set:

${x\in \mathbb{R},x=\frac{n^2-1}{n(n+1)}*\sin{\frac{2^n}{n+3}}-\frac{(-1)^n}{n^2+n+1}*\cos{\frac{(-1)^n*n}{3^n}}, n \in \mathbb{N}}$

I upper-bounded it by it's absolute values and upper bounded the trigonometric functions and by the division of 1 and got the following function:

$\leq n^2+2$

Therefore I determined, that the set has no upper bounds. However, what can I tell about the lower bounds? How do I investigate the lower bounds?

Thanks

William Elliot
  • 17,598
  • Just because you bounded it above by $n^2+2$ doesn't mean it's unbounded above. In fact, since $\dfrac{n^2-1}{n(n-1)} = \dfrac{n+1}{n} = 1 + \dfrac{1}{n}$, the (fairly crude) upper bound I get from the triangle inequality is the following: $$|x| \le 1 + \dfrac{1}{n} + \dfrac{1}{n^2+n+1} \le 1+1+\dfrac{1}{3} = \dfrac{7}{3}$$ – Clive Newstead Feb 18 '20 at 17:50
  • @CliveNewstead And how do I get the lower bound from the upper bound? –  Feb 18 '20 at 17:57
  • Well $|x| \le a$ if and only if $-a \le x \le a$, so this gives you upper and lower bounds. – Clive Newstead Feb 18 '20 at 18:54

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