# Proving sequence converges to 0

1. Mar 17, 2013

### autodidude

1. The problem statement, all variables and given/known data
Prove that the sequence $$a_n=\frac{1}{n}+\frac{(-1)^n}{n^2}$$ converges to 0 using the definition of convergence.

3. The attempt at a solution
I'm pretty stumped on this one...all I've written is $$|\frac{1}{n}+\frac{(-1)^n}{n^2}-0| < \epsilon$$

The only way I know how to attack this problem at the moment is to try and get n in terms of ε but I haven't gotten anywhere with that so far.

2. Mar 17, 2013

### jbunniii

Can you prove that $1/n \rightarrow 0$? If so, try using that along with the fact that $|(-1)^n/n^2| \leq 1/n$.

3. Mar 17, 2013

### autodidude

I think so...

$$|\frac{1}{n}-0|< \epsilon$$
$$\frac{1}{n}< \epsilon$$
$$n>\frac{1}{\epsilon}$$

So if n is greater than 1/ε, then 1/n will be within ε of 0. So we can take N=1/ε and that would mean if we want to be within ε of 0 then take n to be some larger number than N=1/ε (just a bit of thinking aloud there...)

Would we also have to prove $$|(-1)^n/n^2|<1/n$$?

4. Mar 17, 2013

### Fredrik

Staff Emeritus
N is supposed to be an integer, but you can choose it to be any integer that's greater than 1/ε (in the 1/n proof).

If an inequality that you use in the proof isn't obviously true, and you haven't previously proved a theorem that ensures that it holds, you have to prove it. In this case it's rather trivial. (Just simplify it until it becomes obvious).

In your problem, put the problem of choosing an N aside for a while, and just try to show that
$$\left|\frac{1}{n}+\frac{(-1)^n}{n^2}-0\right|$$ is smaller than something that's easier to deal with.

Use \left and \right to make the LaTeX look pretty. (Hit the quote button to see what I did).

5. Mar 17, 2013

### autodidude

Ok I chose 2n (arbitrarily, I'm not aware of any other set criteria for choosing the value) and after a bit of algebra I ended up with $$\left|\frac{(-1)^n}{n}\right|<1$$ which I think is the same as the 'rather trivial' inequality I was asking about.

But so, after choosing a value, where should I go from there?

Last edited: Mar 17, 2013
6. Mar 17, 2013

### Fredrik

Staff Emeritus
I don't understand what you mean by this.

Right, if you take the inequality you asked about in post #3 and cancel n from both sides, this is what you get. But you can certainly simplify it more than this.

Once you have chosen N (if that's what you meant), then you verify that
$$\left|\frac{1}{n}+\frac{(-1)^n}{n^2}-0\right|<\varepsilon.$$ for all integers n such that $n\geq N$. But I still think you should forget about all that until you have shown that
$$\left|\frac{1}{n}+\frac{(-1)^n}{n^2}-0\right|$$ is less than something that's easier to deal with.

Last edited: Mar 17, 2013
7. Mar 18, 2013

### HallsofIvy

Staff Emeritus
Alternatively,
$$\left|\frac{1}{n}- \frac{(-1)^n}{n^2}\right|= \left|\frac{n}{n^2}- \frac{(-1)^n}{n^3}\right|= \left|\frac{n^2+ (-1)^n}{n^2}\right|$$

If n is odd, that is
$$\left|\frac{n^2- 1}{n^2}\right|$$
and if n is even
$$\left|\frac{n^2+ 1}{n^2}\right|$$

so do odd and even n separately.

8. Mar 18, 2013

### Ray Vickson

How you proceed depends on what you already know and what you are allowed to use. For example, do you know (or can use) the result that if both limits on the right exist, then:
$$\lim_{n \to \infty} (f_n + g_n) = \lim_{n \to \infty} f_n + \lim_{n \to \infty} g_n \:?$$
The problem is easy if you can use that result.

9. Mar 18, 2013

### autodidude

You said choose something that's larger than the sequence but easier to deal with...

@HallsOfIvy: Not sure exactly how to work with that at the moment but I'll see what i can do with that approach.

@Ray Vickson: I don't know if I'm allowed to use any limit laws...the question asks says to argue directly from the definition of convergence.

10. Mar 18, 2013

### Dick

Did you try using |a+b|<=|a|+|b|, with a=1/n and b=(-1)^n/n^2? I would have thought so but I'm not sure you did.

11. Mar 18, 2013

### Fredrik

Staff Emeritus
What Dick suggested is what I had in mind as the first step, and I think that's what jbunniii had in mind too.

12. Mar 18, 2013

### Dick

I'm sure it was. Just thought it was about time to make it explicit.