Monotone Sequences

For the nondecreasing sequence, let

The supremum exists because the set is nonempty and bounded above by hypothesis. For each \(\epsilon > 0\), we know that there is some element \(a_N\) of the set which is not much smaller than the supremum, i.e., such that \(a_N > L - \epsilon\). Because the sequence is nondecreasing, for any \(n \geq N\), \(a_n \geq a_N > L - \epsilon\). But by definition of the supremum, we also know that none of the terms are larger than the supremum, i.e., \(a_n \leq L\) as well. Thus \(L - \epsilon < a_n \leq L < L + \epsilon\). Subtracting \(L\) from each part implies that \(|a_n - L| < \epsilon\) for all \(n \geq N\) as desired.

Use the Method of Bisection. Below is an illustration of this method. We have one “guess” \(a_n\) which is known to be too small and a second “guess” \(b_n\) known to be too big. From this pair, we make a new pair of guesses closer together by defining \(a_{n+1}\) and \(b_{n+1}\) so that \([a_{n+1},b_{n+1}]\) is either the left or right half of the interval \([a_n,b_n]\). We make the choice so that the left endpoint is never too large and the right endpoint is never too small. The first figure below illustrates several steps of the method of bisection and the second reiterates how \(a_{n+1}\) and \(b_{n+1}\) are defined in terms of \(a_n\) and \(b_n\).

Let \(A \geq 0\). If we define

We know that \(a_0^k = 0 \leq A\) and \(b_0^k \geq 1 + k A \geq A\), i.e., the \(k\)-th power of \(A\) is too small and the \(k\)-th power of \(b_0\) is too large. The \(k\)-th power of the midpoint \((a_0+b_0)/2\) will either be \(\leq A\) or \(\geq A\). In the former case, let \(a_1 := (a_0+b_0)/2\) and \(b_1 := b_0\); in the latter case let \(a_1 := a_0\) and \(b_1 := (a_0 + b_0)/2\). Repeat this process to define two sequences \(\{a_n\}_{n=0}^\infty\) and \(\{b_n\}_{n=0}^\infty\). Show by induction that

To prove the inequality, observe that

for each value of \(N\) and \(M\), because one can compare both to \(a_n\) for some \(n \geq \max\{N,M\}\). This means that \(\sup_{n \geq M} a_n\) is an upper bound for the set \(\{\inf_{n \geq N} a_n \ : N \geq 1\}\) and so must be greater than or equal to its supremum (if \(\sup_{n \geq M} a_n\) is infinite for all \(M\), then there's nothing to prove because the limsup will be \(+\infty\) which is never strictly smaller than anything; the analogous statement is also true if \(\inf_{n \geq N} a_n = -\infty\) for all \(N\)). Thus

Now repeat; the liminf is a lower bound for the right-hand side, so must be smaller than or equal to the infimum of the right-hand side.

Notice that when \(\limsup_{n \rightarrow \infty} a_n = L\) for some finite \(L\), this means that \(\inf_{N \geq 1} \sup_{n \geq N} a_n = L\). For any positive \(\epsilon\), there must be some \(N\) such that \(\sup_{n \geq N} a_n < L + \epsilon\). That implies, in turn, that \(a_n < L + \epsilon\) for each \(n \geq N\). Similarly, if \(\liminf_{n \rightarrow \infty} a_n = L'\) for some finite \(L'\), then \(a_n > L' - \epsilon\) for all \(n \geq N'\) for some \(N' \in {\mathbb N}\). This is the main observation for proving the relationship to the limit.

Note conversely that if for all \(\epsilon > 0\), there is some \(N\) such that \(a_n < L + \epsilon\), this means that \(\sup_{n \geq N'} a_n \leq L + \epsilon\) for all \(N'\) sufficiently large. This would mean that the infimum also needs to be less than or equal to \(L+ \epsilon\), i.e. it would imply \(\limsup_{n \rightarrow \infty} a_n \leq L + \epsilon\) for each \(L > 0\). That forces \(\limsup_{n \rightarrow \infty} a_n \leq L\). A symmetric argument applies for asymptotic lower bounds and the liminf.