Special Sequences
Most of the previous methods for proving the limit of a sequence require us to have some idea of the limit before we proceed. This section will cover techniques to prove that a limit exists, even if we don’t know the precise value the sequence converges to.
Monotone Sequences
A sequence (s_n) of real numbers is increasing if s_n \leq s_{n+1} for all n \in \mathbb{N}, and is decreasing if s_n \geq s_{n+1} for all n \in \mathbb{N}. A sequence is monotone if it is increasing or decreasing.
A monotone sequence is convergent if and only if it is bounded.
Suppose (s_n) is a bounded increasing sequence. Let S denote the nonempty bounded set \{s_n \colon n \in \mathbb{N}\}. By the completness axiom, S has a least upper bound, and we let s = \sup S. We claim that \lim s_n = s. Given any \varepsilon > 0, s - \varepsilon is not an upper bound for S. Thus there exists a natural number N such that s_N > s - \varepsilon. Since (s_n) is increasing and s is an upper bound for S, we have
s - \varepsilon < s_N \leq s_n \leq s
for all n \geq N. Hence (s_n) converges to s. When the sequence is decreasing, we let s = \inf S and the proof is nearly identical. For completness sake
We claim that \lim s_n = s. Given any \varepsilon > 0, s + \varepsilon is not a lower bound for S. Thus there exists some natural number N such that s_N < s + \varepsilon. Since (s_n) is decreasing and s is a lower bound for S, we have
s + \varepsilon > s_N \geq s_n \geq s
for all n \geq N. Thus (s_n) converges to s.
The converse, that a convergent sequence is bounded, was proved in Theorem 1.1.3.
This allows us to prove sequences are convergent without using the defintion of convergence, and without knowledge of what the sequence converges to.
Let (s_n) be the sequence defined by s_1 = 1 and s_{n+1} = \sqrt{1 + s_n} for n \geq 1.
Claim: (s_n) is a bounded increasing sequence.
We can manually compute the first few terms of the sequences
\begin{align*} s_1 &= 1\\ s_2 &= \sqrt{1 + 1} &\approx 1.414\\ s_3 &= \sqrt{1 + \sqrt{2}} &\approx 1.554\\ s_4 &= \sqrt{1 + \sqrt{1 + \sqrt{2}}} &\approx 1.598 \end{align*}It appears that the sequence is bounded by 2. We can prove this via induction. Suppose s_k < 2 for some k \in \mathbb{N}. Then
s_{k+1} = \sqrt{1 + s_k} < \sqrt{1 + 2} = \sqrt{3} < 2.
Because s_1 = 1, we can conclude via induction that s_n < 2 for all n \in \mathbb{N}. Thus we know (s_n) is bounded. We now want to prove that it’s increasing. We also use induction to prove this. Suppose s_k < s_{k+1} for some natural number k. Then we have
s_{k+1} = \sqrt{1 + s_k} < \sqrt{1 + s_{k+1}} = s_{k+2}
Therefore our induction step is complete. Because s_1 < s_2, we have that s_n < s_{n+1} for all n \in \mathbb{N}. Thus (s_n) is an increasing sequence and is bounded by the interval [1,2]. By Theorem 1.3.1, we can conclude that the sequence is convergent, even though we don’t know the value s the sequence converges to.
Because \lim s_{n+1} = \lim s_n (Proof needed), we know that the value of s must satisfy the equation
s = \sqrt{1 + s}.
This can be reasoned intuitively, as the difference s_{n+1} - s_n must shrink to zero as n grows. Solving through algebra
\begin{align*} s &= \sqrt{1 + s}\\ s^2 - s - 1 &= 0\\ s &= \frac{1 \pm \sqrt{5}}{2} = \phi \end{align*}Where \phi is the Golden ratio. We take the positive value since s_n \geq 1 for all n. Thus \lim s_n = \phi.
(a) If (s_n) is an unbounded increasing sequence, then \lim s_n = + \infty
(b) If (s_n) is an unbounded decreasing sequence, then \lim s_n = - \infty
(a) Let (s_n) be an increasing sequence and suppose that the set S = \{s_n \colon n \in \mathbb{N}\} is unbounded. Since (s_n) is increasing, S is bounded below by s_1. Hence S must be unbounded above. Given any M \in \mathbb{R}, there exists a natural number N such that s_N > M. Since (s_n) is increasing, we have M < s_N \leq s_n, so \lim s_n = + \infty.
Identically
(b) Let (s_n) be a decreasing sequence and suppose that the set S = \{s_n \colon n \in \mathbb{N}\} is unbounded. Since (s_n) is decreasing, S is bounded above by s_1. Thus S must be unbounded below. Given any M \in \mathbb{R}, there exists a natrual number N such that s_N < M. Since (s_n) is decreasing, we have s_n < s_N < M. Therefore \lim s_n = - \infty.
Cauchy sequences
When a sequence converges, not only do the terms get closer to the limit, they get closer to each other for sufficient large n. It turns out that if the terms in a sequence get continually closer to each other, the sequence itself converges.
A sequence (s_n) is said to be a Cauchy sequence if for all \varepsilon > 0, there exists a natural number N such that m,n \geq N implies that |s_n - s_m| < \varepsilon.
Essentially, a sequence is Cauchy if the terms get arbitrarily close to each other.
Every convergent sequence is a Cauchy sequence.
Suppose that (s_n) converges to s. To show that s_n is close to s_m for sufficiently large n,m, we use the fact that they are both close to s. The triangle inequality gives
|s_n - s_m| = |s_n - s + s - s_m| \leq |s_n - s| + |s - s_m|.
For any \varepsilon > 0, choose some natural number N such that k \geq N implies |s_k - s| < \varepsilon/2. This k exists, because \lim s_n = s. Then for m,n \geq N we have
|s_n - s_m| \leq |s_n - s| + |s - s_m| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
Therefore (s_n) is a Cauchy sequence.
Every Cauchy sequence is bounded.
Let (s_n) be a Cauchy sequence. Then there exists a natural number N such that for all n,m \geq N, |s_n - s_m| < \varepsilon. Let \varepsilon = 1. Let m = N. Through the triangle inequality we have
|s_n| - |s_N| \leq |s_n - s_N| < 1 \implies |s_n| < 1 + |s_N|
Let M = \max\{|s_1|,\dots,|s_N|,1 + |s_N|\}. Then for all n \in \mathbb{N}, |s_n| \leq M. Thus (s_n) is bounded.
The above lemma is very similar to Theorem 1.1.3.
A sequence (s_n) is convergent if and only if it is a Cauchy sequence.
We have already proved in Lemma 1.3.3 that every convergent sequence is a Cauchy sequence, so we only need to prove the reverse direction. Suppose (s_n) is a Cauchy sequence and let S = \{s_n \colon n \in \mathbb{N}\} be the range of the sequence. We consider two cases, when S is finite, and when S is infinite.
If S is finite, then the minimum distant \varepsilon between any two points is positive. Since (s_n) is Cauchy, there exists a natural number N such that m,n \geq N implies |s_n - s_m| < \varepsilon. Given any m \geq N, s_m and s_N are both in S. If the distant between them is less than \varepsilon, then it must be zero. Thus s_m = s_N for all m \geq N. Therefore \lim s_n = s_N.
Next, suppose that S is infinite. From Lemma 1.3.4 we know S is bounded. From the Bolzano-Weierstrass theorem there exists a point s in \mathbb{R} that is an accumulation point of S. We claim (s_n) converges to s. Given any \varepsilon > 0, there exists a natural number N such that |s_n - s_m| < \varepsilon/2 for any n,m \geq N. Since s is an accumulation point of S, the neighborhood N(s;\varepsilon/2) contains an infinite amount of points in S. Thus there exists some integer m \geq N such that s_m \in N(s;\varepsilon/2). Hence for any n \geq N we have
|s_n - s| = |s_n - s_m + s_m - s| \leq |s_n - s_m| + |s_m - s| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
Thus \lim s_n = s.
Practice
Recap
In this section we proved the following theorems and results.
A monotone sequence is convergent if and only if it is bounded.
(a) If (s_n) is an unbounded increasing sequence, then \lim s_n = + \infty
(b) If (s_n) is an unbounded decreasing sequence, then \lim s_n = - \infty
Every convergent sequence is a Cauchy sequence.
Every Cauchy sequence is bounded.
A sequence (s_n) is convergent if and only if it is a Cauchy sequence.