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 Rule, 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 following theorems are stated in one order but proved in another order, since we will use the Continuous Function Law to prove the other Laws.

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 \(M\neq 0\).

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 \(M\neq 0\).

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.

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: \[\begin{aligned} |a_n+b_n-L-M|&=|(a_n-L)+(b_n-M)| \\ &\le |a_n-L|+|b_n-M| \lt \dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon \end{aligned}\]

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, \[ |ca_n-cL| = |c||a_n-L| \lt |c|\varepsilon_1=\varepsilon \]

Quotient Law: PY: Proof needed

Power Law: PY: Proof needed

18. Sequences

e3. Precise Limit Laws

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