Limit Properties
We begin the section by showing that algebraic operations are compatible with limits.
Let \lim s_n = s and \lim t_n = t. Then
(a) \; \lim (s_n + t_n) = s + t. \newline (b) \; \lim(ks_n) = ks and \lim(k + s_n) = k + s. \newline (c) \; \lim(s_nt_n) = st. \newline (d) \; \lim(s_n/t_n) = s/t provided \forall n,t_n \not = 0
(a) We must show that |(s_n + t_n) - (s + t)| is arbitrarily close to any \varepsilon. Using the triangle inequality, we have
\begin{align*} \left|(s_n + t_n) - (s + t)\right| &= |(s_n - s) + (t_n - t)|\\ &\leq |s_n - s| + |t_n - t|. \end{align*}
By the definition of convergence we have
\exists N_1 \in \mathbb{N} \text{ such that }\forall n \geq N_1, |s_n - s| < \frac{\varepsilon}{2}.
and
\exists N_2 \in \mathbb{N} \text{ such that }\forall n \geq N_2, |t_n - t| < \frac{\varepsilon}{2}.
Let N = \max\{N_1,N_2\}. Then for all n \geq N, we have |(s_n + t_n) - (s + t)| \leq |s_n - s| + |t_n - t| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
Since this holds for all \varepsilon > 0, we have \lim (s_n + t_n) = s + t.
(b) First, we need to show \lim(ks_n) = ks. Since \lim s_n = s, there exists N \in \mathbb{N} such that for all n \geq N
|s_n - s| \leq \frac{\varepsilon}{|k|}.
Therefore
|k||s_n - s| \leq |k|\frac{\varepsilon}{|k|} \implies |ks_n - s_n| \leq \varepsilon
When k \not = 0. When k = 0, our statement is trivially true. Next we need to show \lim s_n + k = s + k. This is almost trivial since we are given \lim s_n = s, and
|(k + s_n) - (k + s)| = |s_n - s| < \varepsilon.
(c) We must show that |s_nt_n - st| < \varepsilon. We can rewrite our inequality
|s_nt_s - st| = | s_nt_n + (s_nt - s_n t) - st|
We can now use the triangle inequality to show
\begin{align*} |(s_nt_n - s_nt) + (s_n t - st)| &\leq |s_nt_n - s_nt| + | s_nt - st| \\ &= |s_n||t_n - t| + |t||s_n - s| \end{align*}
Since (s_n) is bounded, \exists M_1 > 0 such that \forall n, |s_n| \leq M_1.
Let M = \max\{M_1,|t|\}. Then we have
|s_nt_n - st| \leq M|t_n - t| + M|s_n - s|.
Because \lim s_n = s and \lim t_n = t, for any \varepsilon > 0 there exists N_1,N_2 \in \mathbb{N} such that when n \geq N_1 and n \geq N_2
|t_n - t| \leq \frac{\varepsilon}{2M} \text{ and } |s_n - s| < \frac{\varepsilon}{2M}
respectively. Let N = \max\{N_1,N_2\}. Then n \geq N implies
|s_nt_n - st| \leq M|t_n - t| + M|s_n - s| < M\left(\frac{\varepsilon}{2M}\right) + M\left(\frac{\varepsilon}{2M}\right) = \varepsilon.
Thus we have shown \lim(s_nt_n) = st.
(d) Since s_n/t_n = s_n(1/t_n), if we can prove \lim 1/t_n = 1/t we can use part (c) to complete the proof. We need to show that for any \varepsilon > 0
\left|\frac{1}{t_n} - \frac{1}{t}\right| = \left|\frac{t - t_n}{t_nt}\right| < \varepsilon
For all sufficently large n. We want to get a lower bound on the denominator, t_nt. Because t \not = 0, there exists some N_1 \in \mathbb{N} such that n \geq N_1 implies |t_n - t| < |t|/2 < \varepsilon_t. Thus for n \geq N_1 we have
|t_n| = |t - (t - t_n)| \geq |t| - |t - t_n| > |t| - \frac{|t|}{2} = \frac{|t|}{2}
There also exists N_2 \in \mathbb{N} such that n \geq N_2 implies |t_n - t| < \frac{\varepsilon}{2}|t|^2. Let N = \max\{N_1,N_2\}. Then n \geq N implies
\left|\frac{1}{t_n} - \frac{1}{t}\right| = \left|\frac{t - t_n}{t_nt}\right| = \frac{1}{|t_n|}\left|\frac{t - t_n}{t}\right| < \frac{2}{|t|}\left|\frac{t-t_n}{t}\right|<\varepsilon.
Thus \lim 1/t_n = 1/t.
Another useful property of limits is that they are preserved under inequality operations.
Let (s_n) \to s and (t_n) \to t. If \forall n \in \mathbb{N},\;s_n \leq t_n. Then s \leq t.
Towards a contradiction, suppose that t < s. Let \varepsilon = \frac{s-t}{2}.
By the definition of convergence we have
\exists N_1 \in \mathbb{N} \text{ such that }\forall n \geq N_1, |s_n - s| < \varepsilon.
Similarly we have
\exists N_2 \in \mathbb{N} \text{ such that }\forall n \geq N_2, |t_n - t| < \varepsilon.
Let N = \max\{N_1,N_2\}. For n \geq N, \; t_n < t + \varepsilon = s - \varepsilon < s_n. But this contradicts our inital assumption that s_n \leq t_n for all n. Thus we conclude s \leq t.
This leads to a the following corollary
If (t_n) converges to t and t_n \geq 0 for all n \in \mathbb{N}, then t \geq 0
Let (s_n) be a constant sequence converging to s, such that for all n \in \mathbb{N}, s_n = 0. We are given for all n, t_n \geq 0, thus t_n \geq s_n. By Theorem 1.2.2, we have that t \geq s. Because s = 0, t \geq 0.
These results lead us to the Ratio test. This test is used to show that certain sequences converge to zero.
Suppose that (s_n) is a sequence of positive terms and that the sequence of ratios (s_{n+1}/s_n) converges to L. If L > 1, then \lim_{s_n} = 0
In simple english, if s_n is a sequence of positive terms, where each subsequent term is smaller than the last, the sequence must converge to zero.
By our previous Corollary, we know that L \geq 0. If L < 1, then there exists a real number c such that 0 \leq L < c < 1. Let \varepsilon = c - L, so that \varepsilon > 0.
Since (s_{n+1}/s_n) converges to L, there exists some N \in \mathbb{N} such that for all n \geq N,
\left|\frac{s_{n+1}}{s_n} - L\right| < \varepsilon
Let k = N + 1. Then for all n \geq k we have n - 1 \geq N such that
\frac{s_n}{s_{n+1}} < L + \varepsilon = L + (c - L) = c.
We drop the absolute value signs as, positive or negative, s_{n+1}/s_n - L < \varepsilon. For all n \geq k,
0 < s_n < s_{n-1}c
We can repeat this inequality 0 < s_{n-1} < s_{n-2}c
Thus
0 < s_n < s_{n-1}c < s_{n-2}c^2 < \cdots < s_kc^{n-k}.
Let M = s_k/c^k, we have 0 < s_n < Mc^n for all n \geq k. Since 0 < c < 1, we have \lim c^n = 0 by a previous result. Thus \lim s_n = 0 by Theorem 1.1.1.
Infinite Limits
Certain sequences like s_n = n are clearly not convergent, but behave predictably; growing without bounds towards infinity. We introduce the notion of divergence for these sequences.
A sequence (s_n) is said to diverge to +\infty, notated
\lim s_n = +\infty
If for every M \in \mathbb{R}, there exists a natural number N such that n \geq N implies that s_n > M.
Similarly a sequence (s_n) is said to diverge to -\infty, notated
\lim s_n = -\infty
If for every M \in \mathbb{R} there exists a natural number N such that n \geq N implies that s_n < M.
A divergent sequence is NOT convergent, and our limit notation is only used for simplicity. Theorems which require a convergent sequence do not apply to divergent sequences.
Test
Claim: \lim n^2 = + \infty
To show the sequence (s_n) where s_n = n^2 diverges to positive infinity, we must show for every M \in \mathbb{R}, there exists a natural number N such that n \geq N implies s_n > M.
Let N be equal to the smallest integer greater than |M|. For all integers N^2 \geq N. Thus for all n \geq N, s_n = n^2 > n > M. Thus our sequence (s_n) diverges to positive infinity.
The above example is very simple, and gives a general idea for how infinite limit proofs are done. More technical examples can be found in the practice.
Suppose that (s_n) and (t_n) are sequences such that s_n \leq t_n for all n \in \mathbb{N}.
(a) If \lim s_n = + \infty then \lim t_n = + \infty.
(b) If \lim t_n = - \infty then \lim s_n = - \infty.
(a) We are given that (s_n) diverges to posiitve infinity. Then for every M there exists a naturaul number N such that n \geq N implies s_n > M. Because s_n \leq t_n for all n, we have t_n \geq s_n > M. Thus for any M, there exists some natural number N such that n \geq N implies t_n > M, and we have that \lim t_n = + \infty.
(b) We are given that (t_n) diverges to negative infinity. Then for every M there exists a natural number N such that n \geq N implies t_n < M. Because s_n \leq t_n for all n, we have s_n \leq t_n < M. Thus for every M, there exists some natural number N such that n \geq N implies s_n < M, and we have that \lim s_n = - \infty
Let (s_n) be a sequence of positive numbers. Then \lim s_n = + \infty if and only if \lim 1/s_n = 0.
Suppose that \lim s_n = + \infty. Given any \varepsilon > 0, let M = 1/\varepsilon. There exists a natural number N such that n \geq N implies s_n > M = 1/\varepsilon. Since s_n is positive, we have
\left|\frac{1}{s_n} - 0\right| < \varepsilon = \frac{1}{M}
Thus \lim 1/s_n = 0. For the converse, assume \lim 1/s_n = 0. Then there exists a natural number N such that for all n \geq N, we have
\left|\frac{1}{s_n} - 0\right| = \frac{1}{s_n} < \varepsilon.
Let M = 1/\varepsilon. Flipping both sides of the inequality gives us s_n > 1/\varepsilon = M. Thus for some natural number N, for all n \geq N we have s_n > M. Thus \lim s_n = + \infty.
Practice
Recap
In this section we proved the following theorems and results.
Let \lim s_n = s and \lim t_n = t. Then
(a) \; \lim (s_n + t_n) = s + t. \newline (b) \; \lim(ks_n) = ks and \lim(k + s_n) = k + s. \newline (c) \; \lim(s_nt_n) = st. \newline (d) \; \lim(s_n/t_n) = s/t provided \forall n,t_n \not = 0
Let (s_n) \to s and (t_n) \to t. If \forall n \in \mathbb{N},\;s_n \leq t_n. Then s \leq t.
If (t_n) converges to t and t_n \geq 0 for all n \in \mathbb{N}, then t \geq 0
Suppose that (s_n) is a sequence of positive terms and that the sequence of ratios (s_{n+1}/s_n) converges to L. If L > 1, then \lim_{s_n} = 0
Suppose that (s_n) and (t_n) are sequences such that s_n \leq t_n for all n \in \mathbb{N}.
(a) If \lim s_n = + \infty then \lim t_n = + \infty.
(b) If \lim t_n = - \infty then \lim s_n = - \infty.
Let (s_n) be a sequence of positive numbers. Then \lim s_n = + \infty if and only if \lim 1/s_n = 0.