Direct Consequences

With this language we can give some simple examples for which we can use the definition directly.

• If a_n$\to$2 as n$\to$$\infty$, then (take $\epsilon$ = 1), eventually, a_n is within a distance 1 of 2. One consequence of this is that eventually a_n > 1 and another is that eventually a_n < 3.
• Let a_n = 1/n. Then a_n$\to$ 0 as n$\to$$\infty$. To check this, pick $\epsilon$ > 0 and then choose N with N > 1/$\epsilon$. Now suppose that n$\ge$N. We have
  0$$\le$$$${\frac{{1}}{{n}}}$$$$\le$$$${\frac{{1}}{{N}}}$$ < $$\epsilon$$    by choice of N.    

• The sequence a_n = n - 1 is divergent; for if not, then there is some l such that a_n$\to$l as n$\to$$\infty$. Taking $\epsilon$ = 1, we see that eventually (say after N) , we have -1$\le$(n - 1) - l < 1, and in particular, that (n - 1) - l < 1 for all n$\ge$N. thus n < l + 2 for all n, which is a contradiction.

Exercise 2.5   Show that the sequence a_n = (1/$\sqrt{n}$)$\to$ 0 as n$\to$$\infty$.

Although we can work directly from the definition in these simple cases, most of the time it is too hard. So rather than always working directly, we also use the definition to prove some general tools, and then use the tools to tell us about convergence or divergence. Here is a simple tool (or Proposition).

Proposition 2.6   Let a_n$\to$l as n$\to$$\infty$ and assume also that a_n$\to$m as n$\to$$\infty$. Then l = m. In other words, if a sequence has a limit, it has a unique limit, and we are justified in talking about the limit of a sequence.

Proof. Suppose that l$\ne$m; we argue by contradiction, showing this situation is impossible. Using 1.7, we choose disjoint neighbourhoods of l and m, and note that since the sequence converges, eventually it lies in each of these neighbourhoods; this is the required contradiction. $\qedsymbol$

We can argue this directly (so this is another version of this proof). Pick $\epsilon$ = | l - m|/2. Then eventually | a_n - l| < $\epsilon$, so this holds e.g.. for n$\ge$N₁. Also, eventually | a_n - m| < $\epsilon$, so this holds eg. for n$\ge$N₂. Now let N  =  max(N₁, N₂), and choose n$\ge$N. Then both inequalities hold, and

| l - m| = | l - a_n + a_n - m|

$$\le$$ | l - a_n| + | a_n - m|    by the triangle inequality    

$$\epsilon \epsilon$$ <

Proposition 2.7   Let a_n$\to$l$\ne$ 0 as n$\to$$\infty$. Then eventually a_n$\ne$ 0.

Proof. Remember what this means; we are guaranteed that from some point onwards, we never have a_n = 0. The proof is a variant of ``if a_n$\to$2 as n$\to$$\infty$ then eventually a_n > 1.'' One way is just to repeat that argument in the two cases where l > 0 and then l < 0. But we can do it all in one:

Take $\epsilon$ = | l|/2, and apply the definition of ``a_n$\to$l as n$\to$$\infty$''. Then there is some N such that

$$\ge$$ | a_n - l| <

    Now    l = l - a_n + a_n.

$$\le \le$$     Thus    | l|

$$\ge \ne$$     and    | a_n|

$\qedsymbol$

Exercise 2.8   Let a_n$\to$l$\ne$ 0 as n$\to$$\infty$, and assume that l > 0. Show that eventually a_n > 0. In other words, use the first method suggested above for l > 0.