Prove that a sequence is bounded/unbounded

Question

I'm trying to do a maths problem which requires me to determine whether a sequence is bounded or unbounded and then it wants me to prove my answe.

I know that it's bounded but I've no idea how to prove it. In the previous question in the exercise (where the sequence was unbounded) I found a value for n such that it was greater than the upper bound i.e. N = ceiling function (sqrt(H))ceiling function close +1

(found n in terms of H where H is the upper bound) but I don't know how to do it for this question...

The sequence is:

$$\frac{n(-1)^n - 2^{-n}}{n}$$

Could anyone please help me?

Answer 1

Your sequence is $$a_n=\frac{n(-1)^n-2^{-n}}{n},\qquad n\in\mathbb{N}$$ Or, rewritten to $$a_n=(-1)^n-\frac{1}{n2^n}$$

The first term $(-1)^n$ alternates between 1 and -1, and notice that $\frac{1}{n2^n}$ is always positive, and never greater than one. So it is true that for all $n\in\mathbb{N}$, $-2\leq a_n< 1$, i.e. $(a_n)$ is bounded.

Answer 2

Hint: Try using the triangle inequality to break up the fraction into two terms, and showing each term is bounded.

Answer:

$|\frac{n(-1)^n - 2^{-n}}{n}| \leq |\frac{n(-1)^n}{n}| + |\frac{1}{2^nn}| = 1 +|\frac{1}{2^nn}|\leq 1+1=2$