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. $a_n = \frac{(-1)^{n+1}n + 36}{ \sqrt{n^2 + 2 } }$
  • $\begingroup$ 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
  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 ?
  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.