# Thread: Limit of a sequence

1. ## Limit of a sequence

Hi everyone,

Prove the following:
$\displaystyle \lim_{n \to \infty} (n+2)^2 sin(\frac{1}{n}) = \infty$

I must show that for each $\displaystyle M>0$ we can find $\displaystyle N$ such that $\displaystyle (n+2)^2 sin(\frac{1}{n})> M$ for every $\displaystyle n>N$.

Thing is that I can't figure out how to "get rid of" $\displaystyle sin(\frac{1}{n})$. Is $\displaystyle sin(\frac{1}{n})$ bigger than anything for every $\displaystyle n$?
Thanks!

2. ## Re: Limit of a sequence

For small x, sinx is approximately equal to x. So as n increases sin(1/n) is approximately equal to 1/n

3. ## Re: Limit of a sequence

True, $\displaystyle |sin(\frac{1}{n})|<|\frac{1}{n}|$ but this doesn't really help me. I need to get rid of $\displaystyle sin(\frac{1}{n})$ by showing that it's bigger than something else, not smaller, no?

4. ## Re: Limit of a sequence

I was thinking of replacing sin(1/n) by 1/n and then the limit is clearly infinite, but I suspect that may not do as a solution.

5. ## Re: Limit of a sequence

Yes you can approximate $\displaystyle sin(\frac{1}{n})$ by $\displaystyle \frac{1}{n}$ as n tends to infinity.There is nothing wrong in that.
As an example to enlighten you,even $\displaystyle (n+2)^2$ can be approximated by $\displaystyle n^2$ as n tends to infinity.

6. ## Re: Limit of a sequence

Yes but how does that help me? I need to prove the limit by definition.
For example, I can do this:

$\displaystyle (n+2)^2sin(\frac{1}{n}) > n^2sin(\frac{1}{n}) > nsin(\frac{1}{n}) > ...$

but then what?

7. ## Re: Limit of a sequence

Originally Posted by loui1410
Yes but how does that help me? I need to prove the limit by definition.
For example, I can do this:

$\displaystyle (n+2)^2sin(\frac{1}{n}) > n^2sin(\frac{1}{n}) > nsin(\frac{1}{n}) > ...$

but then what?
$\displaystyle (n+2)^2sin(\frac{1}{n}) > n^2sin(\frac{1}{n}) > n^2(1/n-1/6n^3) = n -1/6n$

8. ## Re: Limit of a sequence

but how did you get to

$\displaystyle ... > n^2sin(\frac{1}{n}) > n^2(\frac{1}{n} - \frac{1}{6n^3}) > ...$ ?

This is the move I'm looking for, but why is this true?

9. ## Re: Limit of a sequence

There are many ways to prove it.But simplest is to look at the Taylor series exapansion of sin(x).
$\displaystyle sin(x)=x-x^3/3!+x^5/5!-....$

10. ## Re: Limit of a sequence

We haven't learned Taylor series yet so I'm not supposed to use it, there must be another solution that they had in mind. Anyway, I'll ask the teaching assistant tomorrow for guidance and see what happens..