Question
Construct a sequence ? of real numbers with the property that for every x ? R there is a subsequence of ? that converges to x.
Explanation / Answer
Sometimes a sequence can be a real function in disguise. In that case the limit of the sequence is the limit of the function, if this limit does exist. Proposition 4.1. Let f(x) be a function of one real variable whose domain contains the interval [1;+1). Let an be the sequence f(n) where n is any natural number. If: lim x!+1 f(x) = L where L is a real number or also 1, then the sequence an has the same limit: lim n!+1 an = L If the function f(x) doesn't have a limit we cannot conclude anything about the sequence: it might or might not converge. Consider for instance an = sin (n). The function f(x) = sin (x) doesn't have a limit but an = sin(n) = 0 for all natural numbers n, so an converges to zero. There are sequences that do not come from a function of one real variable or they come from some non elementary function. For instance pn does not come from a real function if p is negative. The sequence n! comes from a real function, the Gamma function, but unfortunately this function is not an elementary one. In all these cases we might need some specic technique to study the convergence. 2 Proposition 4.2. A sequence an converges to L 2 R if and only if every subsequence converges to L. Example 4.3. The sequence an = (??1)n is not convergent. The subsequence a2n converges to 1 and the subsequence a2n+1 converges to ??1. If an were convergent to some number L all its subsequences should converge to the same L, so an is not convergent. If two subsequences converge to the same limit, in general we cannot conclude that the sequence is convergent. However the subsequence of even members and the subsequence of odd members are special: if they are both convergent and they converge to the same limit, then the original sequence is convergent as well: Proposition 4.4. Let an be a real sequence. If the subsequence a2n converges to a real number L and the subsequence a2n+1 converges to the same number L, then an converges to L as well. Proof. Choose a positive 2 R. Since a2n converges to L, there is p() such that 8n > p(); ja2n ?? Lj < . There is also q() such that 8n > q(); ja2n+1??Lj < . Set n() equal to max(2p(); 2q()+ 1). For all n > n() we are either in the odd or the even case so jan ?? Lj < , which proves the convergence of an to L. This proposition can be used even when the two subsequences go to innity. With the following two propositions we can compare sequences both when they are convergent and divergent. Proposition 4.5 (Squeeze theorem). Let an; bn; cn be three sequences of real numbers. If there is n0 2 N such that 8n n0; an bn cn and both an and cn converge to L 2 R then also bn converges to L. Proof. According to the denition of limit for every > 0 there exists n such that for every n n: L ??