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Prove: $\lim\limits_{n\to \infty} \dfrac{(-1)^n}{n} = 0$ using the definition of a limit:

$\forall_{\epsilon > 0} \exists_N$ st $\forall_{n>N} |a_n - 0| < \epsilon$

  1. I understand the intuition, but I am not sure how to deal with the alternating sign in the sequence and how to deal with the absolute value from the definition of the limit.
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  • $\begingroup$ @Integral: i think you mean $f(x)/g(x)$ $\endgroup$ – Aang Sep 24 '12 at 4:16
  • $\begingroup$ you are right!!! $\endgroup$ – Integral Sep 24 '12 at 15:12
  • $\begingroup$ If $f(x)$ is limited and $g(x)\rightarrow\pm\infty$ when $x\rightarrow\infty$, then $f(x)/g(x)\rightarrow 0$ when $x\rightarrow\infty$. $\endgroup$ – Integral Sep 24 '12 at 15:14
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You need to show that for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $$\left|\frac{(-1)^n}n\right|<\epsilon$$ whenever $n\ge n_\epsilon$. Now $n$ is always positive, so $$\left|\frac{(-1)^n}n\right|=\frac1n\;,$$ and you really just have to show that for each $\epsilon>0$ there is an integer $n_\epsilon$ such that $$\frac1n<\epsilon$$ whenever $n\ge n_0$. If you’re supposed to be doing this rigorously in a real analysis course, you’re probably expected to use the Archimedean property of the reals at this point.

The fact that the terms of the sequence alternate in sign doesn’t really affect anything here: the absolute value kills the sign anyway.

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The absolute value takes care of the alternating sign for you. If I give you an $\epsilon \gt 0$ you need to find an $n$ so that $\left|\frac {(-1)^n}n\right|=\frac 1n \lt \epsilon$. Then argue that if it is true for a given $n$, it is true for all greater numbers.

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What you want to prove is that

For all $\epsilon >0$, there exists a natural number $N$ such that, whenever $n\geq N$, we have $$\left|\frac{(-1)^n}{n}\right|<\epsilon$$

But $\left|\dfrac{(-1)^n}{n}\right|=\dfrac 1 n$.

In fact, what you want to prove is then that

For all $\epsilon >0$, there exists a natural number $N$ such that, whenever $n\geq N$, we have $$\dfrac 1 n <\epsilon$$

This is in fact a consequence of the unboundedness of the natural numbers. Particularily

For all $\epsilon >0$, there exists a natural number $N$ such that, $$\dfrac 1 N <\epsilon$$

is what we call the Archimedean property of the real numbers. Basically, if there was some $\epsilon$ such that for no natural number $n$

$$\dfrac 1 n <\epsilon$$

would hold, this would mean

$$\dfrac 1 n \geq \epsilon$$

for all natural numbers, so

$$\dfrac 1 \epsilon \geq n$$

which would mean $\dfrac 1 \epsilon$ is an upper bound for the natural numbers, which is impossible.

As Ross is suggesting, since $$n<n+1$$ is true for every natural number, you get $$\epsilon >\frac 1 n>\frac 1{n+1}$$ so if there is one $N$ that works, all other naturals greater than it will also work, so you can put all of the above into a proof.

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