# Check if the sequence is bounded?

Before down-voting my question, bare in mind that this is my first post and question here and if you can help me to improve the quality, I'd be really thankful.

I am struggling with how to check if a sequence is bounded when I am checking if it is convergent or divergent. I know how to check whether its decreasing or increasing, find a limit etc.

I googled this a lot and in many particular easy exercises I'm able to check it but I have a few examples where I am totally unable to check it. It is because I do not fully understand how to start and check if the sequence is bounded.

If there is someone who can explain me the process on the following exercises, I would be happy to learn!

1. $$a_n = \frac{1}{n+2}\cdot cos\frac{n \pi}{2}$$
2. $$a_n = (-1)^n\cdot \sum_{k=1}^\infty \frac{1}{k(k+1)}$$
3. $$a_n = \frac{(-1)^{n+1}n + 36}{ \sqrt{n^2 + 2 } }$$
• What have you tried so far? Can you tell us what you've found on Google? – Dr. Mathva Feb 15 at 11:52
• Mr Mathva, I posted a reply below, you can check whats unclear to me. – Exzone Feb 15 at 11:58

1. $$|a_n| \le \frac{1}{n+2}$$ for all $$n$$. Conclusion ?

2. $$\sum_{k=1}^\infty \frac{1}{k(k+1)}=1$$, can you prove this ? (Hint: telescope sum).

Hence $$a_n=(-1)^n$$. Is $$(a_n)$$ bounded ? Is $$(a_n)$$ convergent ?

1. Try to prove: $$a_{2n} \to -1$$ and $$(a_{2n-1} \to 1.$$ Conclusion ?
• For the second, $\sum_{k=1}^\infty \frac{1}{k(k+1)} = (1\cdot(1-\frac{1}{2})+ 1\cdot(\frac{1}{2}-\frac{1}{3})+...+1 \cdot\frac{1}{n}-(\frac{1}{n+1}))$ , then I find the limit when n ->$\infty$which is 1. So it is bounded with one ? Correct? $a_n = (-1)^n$ will always be -1,1,-1,1... so I would say it is bounded with -1 & 1 and the sequence is bounded with -1, 1 ? – Exzone Feb 15 at 12:16

I'll give an example. Let's take your first sequence $$a_n=\frac{1}{n+2}\cos(\frac{n\pi}{2})$$. We need to check if there exists a number $$M>0$$ such that $$|a_n|\leq M$$ for each $$n\in\mathbb{N}$$. Now, it is a very known fact that $$|\cos(x)|\leq 1$$ for any $$x\in\mathbb{R}$$. Also, for any $$n\in\mathbb{N}$$ we have $$\frac{1}{n+2}\leq \frac{1}{1+2}\leq\frac{1}{3}$$. So from here we get that:

$$|a_n|=|\frac{1}{n+2}||\cos(\frac{n\pi}{2})|\leq \frac{1}{3} \forall n\in\mathbb{N}$$

So this sequence is bounded.

• Never mind, I get it now. Thanks for the example. – Exzone Feb 15 at 12:00
• It is a very basic rule: the smaller the denominator is the bigger the fraction is. And vice versa. This is what I used. – Mark Feb 15 at 12:00