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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. $a_n = \frac{(-1)^{n+1}n + 36}{ \sqrt{n^2 + 2 } }$
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  • $\begingroup$ What have you tried so far? Can you tell us what you've found on Google? $\endgroup$ – Dr. Mathva Feb 15 at 11:52
  • $\begingroup$ Mr Mathva, I posted a reply below, you can check whats unclear to me. $\endgroup$ – Exzone Feb 15 at 11:58
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  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 ?
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  • $\begingroup$ 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 ? $\endgroup$ – Exzone Feb 15 at 12:16
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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.

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

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