An Invitation to Pure Mathematics: Accelerated Proofs and Problem Solving 2021

Let \(L=\lim_{n\rightarrow\infty}a_{n}\text{.}\) By the definition of a limit (with \(\epsilon=1\)), there is a natural number \(N\) such that if \(n> N\text{,}\) then \(|a_{n}-L|\lt 1\text{.}\) ¹ Thus, for \(n> N\text{,}\)

For \(n\lt N\text{,}\) we have \(|a_{n}|\leq \max\{|a_i| : 1\leq i\leq N-1\}\text{.}\) Thus, for all \(n\text{,}\)

1. Let \(\epsilon>0\text{.}\) We need to show there is \(N\) so that \(n > N\) implies \(|a_{n}+b_{n}-(a+b)|\lt \epsilon\text{.}\) Note that by the triangle inequality

   \begin{equation*} |a_{n}+b_{n}-(a+b)| =|(a_n-a)+(b_n-b)| \leq |a_{n}-a|+|b_{n}-b| \end{equation*}

   We can make \(|a_{n}+b_{n}-(a+b)|\lt \epsilon\) by making \(|a_{n}-a|\) and \(|b_{n}-b|\) both less than \(\frac{\epsilon}{2}\text{.}\) Since \(a_n\rightarrow a\text{,}\) we know that there is \(N_{1}\) so that \(n > N_1\) implies \(|a_{n}-a|\lt \frac{\epsilon}{2}\text{.}\) Similarly, since \(b_n\rightarrow b\text{,}\) we know that there is \(N_{2}\) so that \(n > N_2\) implies \(|b_{n}-b|\lt \frac{\epsilon}{2}\text{.}\) If we set \(N=\max\{N_1,N_2\}\text{,}\) then for all \(n > N\text{,}\)

   \begin{equation*} |a_{n}+b_{n}-(a+b)| \leq |a_{n}-a|+|b_{n}-b| \lt \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\text{.} \end{equation*}

2. Let \(\epsilon>0\text{.}\) Note that

   \begin{align*} |a_{n}b_{n}-ab| \amp =|a_{n}b_{n}-ab_{n}+ab_{n}-ab| =|(a_{n}-a)b_{n} + a(b_n-b)|\\ \amp \leq |(a_{n}-a)b_{n}| + |a(b_n-b)| =|a_{n}-a|\cdot |b_{n}|+ |a|\cdot |b_{n}-b|\text{.} \end{align*}

   Thus, if we can pick \(N\) large enough so that \(n > N\) implies \(|a_{n}-a|\cdot |b_{n}|\lt \frac{\epsilon}{2}\) and \(|a|\cdot |b_{n}-b|\lt \frac{\epsilon}{2}\text{,}\) then the above will imply \(|a_{n}b_{n}-ab|\lt \epsilon\) and we'll be done. So let's focus on proving these two things. Let's first find \(N_1\) so that \(n > N_1\) implies \(|a_{n}-a|\cdot |b_{n}|\lt \frac{\epsilon}{2}\text{.}\) Since \(b_n\) converges, it is bounded, and so there is \(M\) so that \(|b_{n}|\leq M\) for all \(n\text{,}\) thus

   \begin{equation*} |a_{n}-a|\cdot |b_{n}|\leq M|a_{n}-a|\text{.} \end{equation*}

   Since \(a_n\rightarrow a\text{,}\) there is \(N_1\) so that \(n > N_1\) implies \(|a_{n}-a|\lt \frac{\epsilon}{2M}\) and hence

   \begin{equation*} |a_{n}-a|\leq M|a_{n}-a|\lt M\cdot \frac{\epsilon}{2M}=\frac{\epsilon}{2}\text{.} \end{equation*}

   Now let's find \(N_2\) so that \(n > N_2\) implies \(|a|\cdot |b_{n}-b|\lt \frac{\epsilon}{2}\text{.}\) If \(a=0\text{,}\) then this is always true and we can set \(N_2=1\text{.}\) If \(a\neq 0\text{,}\) then since \(b_n\rightarrow b\text{,}\) there is \(N_2\) so that \(n > N_2\) implies \(|b_{n}-b|\lt \frac{\epsilon}{2|a|}\text{,}\) and so that

   \begin{equation*} |a|\cdot |b_{n}-b|\lt |a|\cdot \frac{\epsilon}{2|a|}=\frac{\epsilon}{2}\text{.} \end{equation*}

   Thus, if we set \(N=\max\{N_{1},N_{2}\}\text{,}\) then if \(n > N\text{,}\) then

   \begin{equation*} |a_{n}b_{n}-ab| \leq |a_{n}-a|\cdot |b_{n}|+ |a|\cdot |b_{n}-b| \lt \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\text{.} \end{equation*}

3. We leave this one as an exercise.

4. We claim that it suffices to show that if \(b_n\rightarrow b\) and \(b_n,b\neq 0\text{,}\) then \(b_{n}^{-1}\rightarrow b^{-1}\text{.}\) Indeed, if this is the case, then by part (b),

   \begin{equation*} \frac{a_{n}}{b_{n}}=a_{n}\cdot b_{n}^{-1} \rightarrow ab^{-1}=\frac{a}{b}\text{.} \end{equation*}

   So now let's prove that if \(b_n\rightarrow b\) and \(b_n,b\neq 0\text{,}\) then \(b_{n}^{-1}\rightarrow b^{-1}\text{.}\) Let \(\epsilon>0\text{.}\) We need to find \(N\) so that \(n > N\) implies \(|b_{n}^{-1}-b^{-1}|\lt \epsilon\text{.}\) Observe that

   \begin{equation*} \left|\frac{1}{b_{n}}-\frac{1}{b}\right| =\left|\frac{b-b_{n}}{bb_{n}}\right| =\frac{|b-b_{n}|}{|b|\cdot |b_{n}|}\text{.} \end{equation*}

   Suppose we can show that for some number \(s>0\) we have

   \begin{equation} |b_{n}|\geq s \text{ for every } n\in\mathbb{N}\text{.}\label{e_bn_m}\tag{5.2.1} \end{equation}

   Then since \(b_{n}\rightarrow b\text{,}\) we know that for any \(\epsilon_1>0\) there is \(N = N(\epsilon_1)\) so that \(n > N\) implies \(|b_{n}-b|\lt \epsilon_1\text{,}\) and then we would have

   \begin{equation*} \left|\frac{1}{b_{n}}-\frac{1}{b}\right|\leq \frac{|b-b_{n}|}{|b|\cdot |b_{n}|} \leq \frac{\epsilon_1}{|b| s}\text{.} \end{equation*}

   Hence, if we pick \(\epsilon_1=|b|s\epsilon\text{,}\) then all this implies that for \(n > N(|b|s\epsilon)\text{,}\)

   \begin{equation*} \left|\frac{1}{b_{n}}-\frac{1}{b}\right| \leq \frac{\epsilon_1}{|b| s}= \frac{|b|s\epsilon}{|b| s}=\epsilon\text{.} \end{equation*}

   So now we focus on proving (5.2.1). Since \(b_n\rightarrow b\text{,}\) there is \(N'\) so that \(n > N'\) implies \(|b_{n}-b|\lt \frac{|b|}{2}\text{.}\) Thus, by the reverse triangle inequality,

   \begin{equation*} |b_{n}| =|b_{n}-b+b| \geq |b|-|b_{n}-b| >|b|-\frac{|b|}{2} = \frac{|b|}{2}\text{.} \end{equation*}

   For \(n \leq N'\text{,}\) we simply have \(|b_{n}|\geq \min\{|b_{1}|,\dots,|b_{N'}|\}>0\text{.}\) Thus, for all \(n\text{,}\)

   \begin{equation*} |b_{n}|\geq \min\left\{\frac{|b|}{2},|b_{1}|,\dots ,|b_{N'}|\right\}\text{.} \end{equation*}

   This finishes the proof of (5.2.1), and thus of (d).

We prove this by induction on \(k\text{.}\) The base case \(k=1\) is a tautology. For the induction step, assume we have shown that whenever \(x_n\rightarrow L\text{,}\) then for some particular \(k \geq 1\text{,}\) \(x_{n}^{k}\rightarrow L^{k}\text{.}\) Now apply Proposition 5.2.3(b) with \(a_n=x_{n}^{k}\) and \(b_{n}=x_{n}\text{,}\) to obtain

Thus, the proposition follows by induction.

1. Suppose for a contradiction that \(x>y\text{.}\) Take \(\epsilon = (x-y)/2 > 0\text{.}\) Since \(x_{n}\rightarrow x\text{,}\) there is \(N_{1}\) so that \(n > N_{1}\) implies

   \begin{equation*} x-\frac{x-y}{2}\lt x_{n}\lt x+\frac{x-y}{2} \end{equation*}

   and in particular

   \begin{equation*} \frac{x+y}{2}\lt x_{n}\text{.} \end{equation*}

   Similarly, since \(y_{n}\rightarrow y\text{,}\) there is \(N_{2}\) so that \(n > N_{2}\) implies

   \begin{equation*} y-\frac{x-y}{2}\lt y_{n}\lt y+\frac{x-y}{2} \end{equation*}

   and in particular

   \begin{equation*} y_{n}\lt \frac{x+y}{2}\text{.} \end{equation*}

   Therefore, for all \(n> \max\{N_{1},N_{2}\}\text{,}\)

   \begin{equation*} \frac{x+y}{2}\lt x_{n} \leq y_n \lt \frac{x+y}{2}\text{.} \end{equation*}

   which gives the desired contradiction.

2. This follows by (a) by letting \(y_{n}=y\) for all \(n\text{.}\)

Let \(\epsilon > 0\text{.}\) There exist \(N_1\) and \(N_2\) such that for \(n > N_1\) we have

and for \(n > N_2\) we have

Take \(N = \max\{N_1, N_2\}\text{.}\) For \(n > N\) we have

Therefore for \(n > N\) we have \(|y_n - L| \lt \epsilon\text{.}\)