18. Sequences

\(\displaystyle \lim_{n\to\infty}a_n=L\) means For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|a_n-L| \lt \varepsilon\). \(\displaystyle \lim_{n\to\infty}a_n=\infty\) means For all \(M \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(a_n \gt M\).

e3. Precise Limit Laws

Many of the proofs will need the following two results:

\[ \left|\dfrac{}{}|X|-|Y|\dfrac{}{}\right| \le |X+Y| \le |X|+|Y| \]

Triangle Inequality:
If \(X\) and \(Y\) are both positive or both negative, then \(|X+Y|=|X|+|Y|\), which is the equality case of the right side. Further, \(|X+Y|\) is greater than both \(|X|-|Y|\) and \(|Y|-|X|\) which is what the left side says.

Suppose \(X\) and \(Y\) have opposite signs. Let \(X\) be positive and \(Y\) be negative. Then \(Y=-Z\) where \(Z\) is positive. Consequently, \(|X+Y|=|X-Z| \le |X|+|Z|=|X|+|Y|\), which is the right inequality. Further, \(|X+Y|=|X-Z|=||X|-|Y||\), which is the equality case of the left side.

\[ \lim_{n\to\infty}a_n=L \qquad \text{if and only if} \qquad \lim_{n\to\infty}(a_n-L)=0 \]

Equivalent Limit Lemma:
On the one hand, \(\displaystyle \lim_{n\to\infty}a_n=L\) means For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|a_n-L| \lt \varepsilon\). On the other hand, \(\displaystyle \lim_{n\to\infty}(a_n-L)=0\) means For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|(a_n-L)-0| \lt \varepsilon\). They are the same statement.

We can now prove the first two Limit Laws:

Addition Law: The conditions \(\displaystyle \lim_{n\to\infty}a_n=L\) and \(\displaystyle \lim_{n\to\infty}b_n=M\) say: For all \(\varepsilon_1 \gt 0\), there is a positive integer, \(N_1\), such that if \(n \gt N_1\) then \(|a_n-L| \lt \varepsilon_1\).
For all \(\varepsilon_2 \gt 0\), there is a positive integer, \(N_2\), such that if \(n \gt N_2\) then \(|b_n-M| \lt \varepsilon_2\). We need to show \(\displaystyle \lim_{n\to\infty}(a_n+b_n)=L+M\) which means: For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|a_n+b_n-L-M| \lt \varepsilon\). So given an arbitrary number \(\varepsilon \gt 0\), we let \[ \varepsilon_1=\varepsilon_2=\dfrac{\varepsilon}{2} \] and pick \(N_1\) and \(N_2\) to satisfy the definitions of the limits of \(a_n\) and \(b_n\). Then, let \(N=\max(N_1,N_2)\). Then, if \(n \gt N\) then \(n \gt N_1\) and \(n \gt N_2\). Consequently, \[ |a_n-L| \lt \varepsilon_1=\dfrac{\varepsilon}{2} \quad \text{and} \quad |b_n-M| \lt \varepsilon_2=\dfrac{\varepsilon}{2} \] Finally, by the Triangle Inequality: \[\begin{aligned} |a_n+b_n-L-M|&=|(a_n-L)+(b_n-M)| \\ &\le |a_n-L|+|b_n-M| \lt \dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon \end{aligned}\]

Subtraction Law: Let \(c_n=-b_n\). Then by the Constant Multiple Law (proved below), \(\displaystyle \lim_{n\to\infty}c_n=-\lim_{n\to\infty}b_n=-M\). And by the addition law, \(\displaystyle \lim_{n\to\infty}(a_n-b_n) =\lim_{n\to\infty}(a_n+c_n)=L-M\).

Before proving the Product Law, we prove two special cases which will be helpfiul in its proof:

Let \(a_n\) be a sequence for which \(\displaystyle \lim_{n\to\infty}a_n=L\) where \(L\) is finite. Then for all constants \(c\), we have: \[ \lim_{n\to\infty}c\,a_n=cL \]

Constant Multiple Law: The condition \(\displaystyle \lim_{n\to\infty}a_n=L\) says: For all \(\varepsilon_1 \gt 0\), there is a positive integer, \(N_1\), such that if \(n \gt N_1\) then \(|a_n-L| \lt \varepsilon_1\). We need to show \(\displaystyle \lim_{n\to\infty}(ca_n)=cL\) which means: For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|ca_n-cL| \lt \varepsilon\). So given an arbitrary number \(\varepsilon \gt 0\), we let \[ \varepsilon_1=\dfrac{\varepsilon}{|c|} \] and pick \(N_1\) to satisfy the definition of the limit of \(a_n\), and let \(N=N_1\). Then, if \(n \gt N\) then \(n \gt N_1\) also. Consequently, \[ |ca_n-cL| = |c||a_n-L| \lt |c|\varepsilon_1=\varepsilon \]

Let \(a_n\) and \(b_n\) be sequences for which \(\displaystyle \lim_{n\to\infty}a_n=0\) and \(\displaystyle \lim_{n\to\infty}b_n=0\). Then: \[ \lim_{n\to\infty}a_n\,b_n=0 \]

Product Law for Zero Limits: The conditions \(\displaystyle \lim_{n\to\infty}a_n=0\) and \(\displaystyle \lim_{n\to\infty}b_n=0\) say: For all \(\varepsilon_1 \gt 0\), there is a positive integer, \(N_1\), such that if \(n \gt N_1\) then \(|a_n| \lt \varepsilon_1\).
For all \(\varepsilon_2 \gt 0\), there is a positive integer, \(N_2\), such that if \(n \gt N_2\) then \(|b_n| \lt \varepsilon_2\). We need to show \(\displaystyle \lim_{n\to\infty}(a_nb_n)=0\) which means: For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|a_nb_n| \lt \varepsilon\). So given an arbitrary number \(\varepsilon \gt 0\), we let \[ \varepsilon_1=\varepsilon_2=\sqrt{\varepsilon} \] and pick \(N_1\) and \(N_2\) to satisfy the definitions of the limits of \(a_n\) and \(b_n\). Then, let \(N=\max(N_1,N_2)\). Then, if \(n \gt N\) then \(n \gt N_1\) and \(n \gt N_2\). Consequently, \[ |a_n| \lt \varepsilon_1=\sqrt{\varepsilon} \quad \text{and} \quad |b_n| \lt \varepsilon_2=\sqrt{\varepsilon} \] Consequently: \[ |a_nb_n| \le \varepsilon_1\varepsilon_2=\sqrt{\varepsilon}\sqrt{\varepsilon} =\varepsilon \]

We can now prove the Product Rule:

Product Law: By the Equivalent Limit Lemma, the theorem statement is equivalent to:
Let \(a_n\) and \(b_n\) be sequences for which \(\displaystyle \lim_{n\to\infty}(a_n-L)=0\) and \(\displaystyle \lim_{n\to\infty}(b_n-M)=0\) where \(L\) and \(M\) are finite. Then: \[ \lim_{n\to\infty}(a_n\,b_n-LM)=0 \] The proof of this alternate statement is a straightforward computation:

\[\begin{aligned} \lim_{n\to\infty}&(a_n\,b_n-LM) \\[5pt] &=\lim_{n\to\infty}([(a_n-L)+L][(b_n-M)+M]-LM) \\[8pt] &=\lim_{n\to\infty}[(a_n-L)(b_n-M)+M(a_n-L)+L(b_n-M)] \\[8pt] &=\lim_{n\to\infty}(a_n-L)(b_n-M) +\lim_{n\to\infty}M(a_n-L)+\lim_{n\to\infty}L(b_n-M) \\[8pt] &=\lim_{n\to\infty}(a_n-L)(b_n-M) +M\lim_{n\to\infty}(a_n-L)+L\lim_{n\to\infty}(b_n-M) \\[8pt] &=0+M0+L0 \\[2pt] &=0 \end{aligned}\]

Subtract and add \(L\) and \(M\).\(\dfrac{}{\dfrac{}{}}\)
Distribute (FOIL).\(\dfrac{}{\dfrac{}{}}\)
Sum Law.\(\dfrac{}{\dfrac{}{}}\)
Constant Multiple Law.\(\dfrac{}{\dfrac{}{}}\)
\(\displaystyle \lim_{n\to\infty}(a_n-L)=0\), \(\displaystyle \lim_{n\to\infty}(b_n-M)=0\)
and Product Law for Zero Limits

The Continuous Function Law is proved next beacuse it is needed for the proofs of the Reciprocal and Power Laws.

Let \(a_n\) be a sequence for which \(\displaystyle \lim_{n\to\infty}a_n=L\) where \(L\) is finite and let \(f(x)\) be a function which is continuous at \(L\) and at each of the numbers \(a_n\). Then: \[ \lim_{n\to\infty}f(a_n)=f(L) \]

Continuous Function Law: The condition \(\displaystyle \lim_{n\to\infty}a_n=L\) say: For all \(\varepsilon_1 \gt 0\), there is a positive integer, \(N_1\), such that if \(n \gt N_1\) then \(|a_n-L| \lt \varepsilon_1\). The fact that \(f(x)\) is continuous at \(x=L\) say: For all \(\varepsilon_2 \gt 0\), there is a \(\delta_2 \gt 0\), such that if \(0 \lt |x-L| \lt \delta_2\) then \(|f(x)-f(L)| \lt \varepsilon_2\). We need to show \(\displaystyle \lim_{n\to\infty}(f(a_n))=f(L)\) which means: For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|f(a_n)-f(L)| \lt \varepsilon\). So given an arbitrary number \(\varepsilon \gt 0\), we let \[ \varepsilon_1=\varepsilon_2=\dfrac{\varepsilon}{2} \] and pick \(N_1\) and \(N_2\) to satisfy the definitions of the limits of \(a_n\) and \(b_n\). Then, let \(N=\max(N_1,N_2)\). Then, if \(n \gt N\) then \(n \gt N_1\) and \(n \gt N_2\). Consequently, \[ |a_n-L| \lt \varepsilon_1=\dfrac{\varepsilon}{2} \quad \text{and} \quad |b_n-M| \lt \varepsilon_2=\dfrac{\varepsilon}{2} \] Finally, by the Triangle Inequality: \[\begin{aligned} |a_n+b_n-L-M|&=|(a_n-L)+(b_n-M)| \\ &\le |a_n-L|+|b_n-M| \lt \dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon \end{aligned}\]

The Quotient Law follows from the Product Rule and the following special case:

Let \(b_n\) be a sequence for which \(\displaystyle \lim_{n\to\infty}b_n=M\) where \(M\) is finite. Then: \[ \lim_{n\to\infty}\dfrac{1}{b_n}=\dfrac{1}{M} \] provided \(b_n\neq 0\) for all \(n\) and \(M\neq 0\).

PY: Proof needed. This if for the wrong rule.
Reciprocal Law: The condition \(\displaystyle \lim_{n\to\infty}a_n=L\) says: For all \(\varepsilon_1 \gt 0\), there is a positive integer, \(N_1\), such that if \(n \gt N_1\) then \(|a_n-L| \lt \varepsilon_1\). We need to show \(\displaystyle \lim_{n\to\infty}(ca_n)=cL\) which means: For all \(\varepsilon \gt 0\), there is a positive integer, \(N\), such that if \(n \gt N\) then \(|ca_n-cL| \lt \varepsilon\). So given an arbitrary number \(\varepsilon \gt 0\), we let \[ \varepsilon_1=\dfrac{\varepsilon}{|c|} \] and pick \(N_1\) to satisfy the definition of the limit of \(a_n\), and let \(N=N_1\). Then, if \(n \gt N\) then \(n \gt N_1\) also. Consequently, \[ |ca_n-cL| = |c||a_n-L| \lt |c|\varepsilon_1=\varepsilon \]

Let \(a_n\) and \(b_n\) be sequences for which \(\displaystyle \lim_{n\to\infty}a_n=L\) and \(\displaystyle \lim_{n\to\infty}b_n=M\) where \(L\) and \(M\) are finite. Then: \[ \lim_{n\to\infty}\dfrac{a_n}{b_n}=\dfrac{L}{M} \] provided \(b_n\neq 0\) for all \(n\) and \(M\neq 0\).

Quotient Law: PY: Proof needed

Let \(a_n\) and \(b_n\) be sequences for which \(\displaystyle \lim_{n\to\infty}a_n=L\) and \(\displaystyle \lim_{n\to\infty}b_n=M\) where \(L\) and \(M\) are finite. Then: \[ \lim_{n\to\infty}(b_n)^{a_n}=M^L \] provided \(b_n\gt0\) for all \(n\), \(M\gt0\), \((b_n)^{a_n }\) is defined for all \(n\) and \(M^L\) is defined.

Power Law: PY: Proof needed

PY: Need precise proofs of Reciprocal, Quotient and Power Laws

18. Sequences

e3. Precise Limit Laws

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