A.2 Chapter 2

(i)

The sequence $(q_{n})_{n\in\mathbb{N}}$ where $q_{n}\in\mathbb{N}_{0}$ is the unique value for which there exists some $r\in\{0,1,2,3,4\}$ such that $n=5q_{n}+r$.

(ii)

The sequence $(r_{n})_{n\in\mathbb{N}}$, where $r_{n}\in\{0,1,2,3,4\}$ is the unique value for which there exists some $q\in\mathbb{N}_{0}$ such that $n=5q+r_{n}$.

(iii)

The sequence $\big{(}\frac{r_{n}}{q_{n}+1}\big{)}_{n\in\mathbb{N}}$ where $q_{n}\in\mathbb{N}_{0}$ and $r_{n}\in\{0,1,2,3,4\}$ are defined by $n=5q_{n}+r_{n}$ for all $n\in\mathbb{N}$.

The sequence $(a_{n})_{n\in\mathbb{N}}$ where $a_{n}:=1-1/n$ for $n$ even and $a_{n}:=1/(n+1)$ for $n$ odd.

$(1$, $1$, $2$, $3$, $5$, $8$, $13$, $21,\dots)$.

The sequence is defined recursively by ‘saying out loud’ the digits of the previous term. In particular we have

and so on. Since $312211$ is ‘one three, one one, two twos, two ones’, the next term of the sequence is $13112221$.

Suppose $(a_{n})_{n\in\mathbb{N}}$ is a non-increasing sequence. Let $n\in\mathbb{N}$ so that, by hypothesis $a_{n+1}\leq a_{n}$ and therefore $-a_{n+1}\geq-a_{n}$. Since this holds for all $n\in\mathbb{N}$, we conclude that $(-a_{n})_{n\in\mathbb{N}}$ is a non-decreasing sequence.

(i) For all $n\in\mathbb{N}$, we have

since $n^{2}\geq n$ for $n\geq 1$. Thus,

and therefore $3$ is an upper bound for $\big{(}\tfrac{n+2}{2n^{2}-n}\big{)}_{n\in\mathbb{N}}$. Since an upper bound exists, the sequence is bounded above.

(ii) For all $n\in\mathbb{N}$, we have

and therefore

Let $M\in\mathbb{R}$ and choose $n\in\mathbb{N}$ with $n>M$. The above inequality shows that $a_{n}>M$ and so $M$ is not an upper bound for the sequence $(a_{n})_{n\in\mathbb{N}}$. Since $M$ was chosen arbitrarily, we conclude that this sequence is not bounded above.

Suppose $(a_{n})_{n\in\mathbb{N}}$ is bounded. Then there exists $M\in\mathbb{R}$ and $L\in\mathbb{R}$ such that $L\leq a_{n}\leq M$ for all $n\in\mathbb{N}$. If we choose $R:=\max\{|L|,|M|\}$, then it follows that

and therefore $|a_{n}|\leq R$ for all $n\in\mathbb{N}$.

Conversely, suppose there exists some $R>0$ such that $|a_{n}|\leq R$ for all $n\in\mathbb{N}$. Taking $M:=R$ and $L:=-R$, we have $L\leq a_{n}\leq M$ for all $n\in\mathbb{N}$ and so $(a_{n})_{n\in\mathbb{N}}$ is bounded.

Suppose $(a_{n})_{n\in\mathbb{N}}$ is bounded below. Then there exists some $L\in\mathbb{R}$ such that $a_{n}\geq L$ for all $n\in\mathbb{N}$. Defining $M:=-L$, it follows that $-a_{n}\leq-L=M$ for all $n\in\mathbb{N}$ and so $(-a_{n})_{n\in\mathbb{N}}$ is bounded above.

(i) We claim that $\tfrac{3n}{2n+5}\to\frac{3}{2}$ as $n\to\infty$.

For $n\in\mathbb{N}$, we have

where in the last step we used the fact that $2(2n+5)>4n$ for all $n\in\mathbb{N}$.

Let $\varepsilon>0$ be given and choose $N:=\lceil 4/\varepsilon\rceil$. If $n\in\mathbb{N}$ satisfies $n>N$, then it follows from our earlier work that

Hence, by the $\varepsilon$-$N$ definition of a limit, $\tfrac{3n}{2n+5}\to\frac{3}{2}$ as $n\to\infty$.

(ii) We claim that $\tfrac{n}{n^{2}+1}\to 0$ as $n\to\infty$.

For $n\in\mathbb{N}$, we have

Let $\varepsilon>0$ be given and choose $N:=\lceil 1/\varepsilon\rceil$. If $n\in\mathbb{N}$ satisfies $n>N$, then

Hence, by the $\varepsilon$-$N$ definition of a limit, $\tfrac{n}{n^{2}+1}\to 0$ as $n\to\infty$.

Let $\varepsilon>0$ be given and choose $N:=1$. Then for all $n\in\mathbb{N}$ with $n>N$, we have

Thus, by the $\varepsilon$-$N$ definition of a limit, $a_{n}\to a$ as $n\to\infty$.

Given $0<\rho<1$, we may write

On the other hand, given $n\in\mathbb{N}$, the Bernoulli inequality tells us that $(1+\rho)^{n}\geq 1+n\rho$. Combining these observations,

Now let $0<r<1$ and write $r=1-\rho$ for some $0<\rho<1$. Let $\varepsilon>0$ be given and choose $N:=\lceil 1/(\rho\varepsilon)\rceil$. If $n\in\mathbb{N}$ satisfies $n>N$, then

Thus, by the $\varepsilon$-$N$ definition of a limit, $r^{n}\to 0$ as $n\to\infty$.

It is clear that $a_{1}=0=1-1=1-10^{0}$. For $n\in\mathbb{N}$ with $n\geq 2$, we have

Hence, $a_{n}=1-10^{-(n-1)}$ for all $n\in\mathbb{N}$.

Let $\varepsilon>0$ be given. From Exercise 2.31, we know that $10^{-n}\to 0$ as $n\to\infty$ and so by the $\varepsilon$-$N$ definition of a limit there exists some $N\in\mathbb{N}$ such that $10^{-n}<\varepsilon/10$ for all $n\in\mathbb{N}$ with $n>N$. Consequently, if $n\in\mathbb{N}$ satisfies $n>N$, then

Hence, by the $\varepsilon$-$N$ definition of a limit, $a_{n}\to 1$ as $n\to\infty$.

For $n\in\mathbb{N}$, we have

Let $\varepsilon>0$ be given and choose $N:=\lceil 5/\varepsilon\rceil\in\mathbb{N}$. If $n\in\mathbb{N}$ with $n>N$, then

Thus, by the $\varepsilon$-$N$ definition of a limit, $\sqrt{n^{2}+5}-n\to 0$ as $n\to\infty$.

1. (i)
   ​

   (a).

2. (ii)
   ​

   (c).

3. (iii)
   ​

   (b);

4. (iv)
   ​

   (b);

5. (v)
   ​

   (b);

6. (vi)
   ​

   (d).

The justification is as follows:

(i) $\iff$ (a). This is the $\varepsilon$-$N$ definition of a limit.

(ii) $\iff$ (c). Suppose (ii) holds. Then taking $N:=1$, for all $n\in\mathbb{N}$ we have $|a_{n}-a|<\varepsilon$ for all $\varepsilon>0$. However, this can only hold if $|a_{n}-a|=0$ so that $a_{n}=a$ for all $n\in\mathbb{N}$. Thus (c) holds.

Conversely, suppose (c) holds so that $a_{n}=a$ for all $n\in\mathbb{N}$. Then for all $\varepsilon>0$ we have $|a_{n}-a|=0<\varepsilon$ for all $n\in\mathbb{N}$, so (ii) holds.

(iii) $\iff$ (b). Suppose (iii) holds. Taking $N=1$, there exists some $\varepsilon>0$ such that $|a_{n}-a|<\varepsilon$ for all $n\in\mathbb{N}$. Thus, by the triangle inequality, $|a_{n}|\leq|a|+|a_{n}-a|<|a|+\varepsilon$ for all $n\in\mathbb{N}$. Thus, $|a_{n}|\leq M:=|a|+\varepsilon$ for all $n\in\mathbb{N}$, so (b) holds.

Conversely, suppose (b) holds. Then there exists some $M>0$ such that $|a_{n}|<M$ for all $n\in\mathbb{N}$. Let $\varepsilon:=M+|a|$. Then $|a_{n}-a|\leq|a_{n}|+|a|<\varepsilon$ for all $n\in\mathbb{N}$. Thus, (iii) holds.

(iv) $\iff$ (b). Suppose (iv) holds, so there exists some $N\in\mathbb{N}$ and some $\varepsilon>0$ such that $|a_{n}-a|<\varepsilon$ for all $n>N$. Thus, by the triangle inequality, $|a_{n}|\leq|a|+|a_{n}-a|<|a|+\varepsilon$ for all $n>N$. From this we conclude that

so that (b) holds.

Conversely, suppose (b) holds. Then there exists some $M>0$ such that $|a_{n}|<M$ for all $n\in\mathbb{N}$. Let $\varepsilon:=M+|a|$. Then $|a_{n}-a|\leq|a_{n}|+|a|<\varepsilon$ for all $n\in\mathbb{N}$. Thus, (iii) holds.

(v) $\iff$ (b). Suppose (v) holds. Taking $N=1$, there exists some $\varepsilon>0$ such that $|a_{n}-a|<\varepsilon$ for all $n\in\mathbb{N}$. Thus, $|a_{n}|\leq M:=|a|+\varepsilon$ for all $n\in\mathbb{N}$, so that (b) holds.

Conversely, suppose (b) holds. Then there exists some $M>0$ such that $|a_{n}|<M$ for all $n\in\mathbb{N}$. Let $\varepsilon:=M+|a|$. Then $|a_{n}-a|\leq|a_{n}|+|a|<\varepsilon$ for all $n\in\mathbb{N}$. Thus, (v) holds.

(vi) $\iff$ (d). The condition $|a_{n}-a|<\varepsilon$ for all $\varepsilon>0$ holds if and only if $|a_{n}-a|=0$, which in turn holds if and only if $a_{n}=a$. Thus, (vi) holds if and only if there exists some $N\in\mathbb{N}$ such that $a_{n}=a$ for all $n>N$, which is precisely the condition (d).

(i) We claim that $\frac{10n^{3}+5n+1}{n^{3}+3n^{2}+5}\to 10$ as $n\to\infty$.

Multiplying both the numerator and denominator by $1/n^{3}$ gives

Since we know $1/n\to 0$ as $n\to\infty$, it follows from parts 1 and 2 of Theorem 2.38 that

Since the limit of the denominator is non-zero, we can apply part 3 of Theorem 2.38 to conclude that

as required.

(ii) We claim that $\tfrac{2^{n}+3}{2^{n+1}-2}\to\tfrac{1}{2}$ as $n\to\infty$.

Multiplying both the numerator and denominator by $2^{-n}$ gives

Since we know from Exercise 2.31 that $2^{-n}\to 0$ as $n\to\infty$, it follows from parts 1 and 2 of Theorem 2.38 that

Since the limit of the denominator is non-zero, we can apply part 3 of Theorem 2.38 to conclude that

as required.

For $n\in\mathbb{N}$, observe that

Since $b_{n}\to b$ as $n\to\infty$, we expect to be able to replace the $b_{n}$ on the denominator with $b$. More precisely, by the $\varepsilon$-$N$ definition of a limit, since $b\neq 0$ there exists some $N_{1}\in\mathbb{N}$ such that

If $n\in\mathbb{N}$ satisfies $n>N_{1}$, then it follows by the triangle inequality that

and so

Let $\varepsilon>0$. Since $b_{n}\to b$ as $n\to\infty$ and $b\neq 0$, by the $\varepsilon$-$N$ definition of a limit, there exists some $N_{2}\in\mathbb{N}$ such that

If we define $N:=\max\{N_{1},N_{2}\}$, then

If $n\in\mathbb{N}$ satisfies $n>N_{1}$, then it follows from our earlier observations that

Hence, by the $\varepsilon$-$N$ definition of a limit, $1/b_{n}\to 1/b$ as $n\to\infty$.

For $n\in\mathbb{N}$, we have

We know from Exercise 2.31 that $2^{-n}\to 0$ as $n\to\infty$ and, by the limit laws, $-2^{-n}\to 0$ as $n\to\infty$. Hence, by the squeeze theorem, $\frac{(-1)^{n}}{2^{n}+5}\to 0$ as $n\to\infty$.

Let $A\subseteq\mathbb{R}$ be a nonempty set which is bounded above, so that $s:=\sup A$ exists by the completeness axiom.

(i) Given $n\in\mathbb{N}$, using the approximation property for suprema from Lemma 1.31 with $\varepsilon:=1/n$, there exists some $a_{n}\in A$ such that $s-1/n<a_{n}\leq s$.

(ii) From part (i) we have $\ell_{n}\leq a_{n}\leq u_{n}$ where $\ell_{n}:=s-1/n$ and $u_{n}:=s$ for all $n\in\mathbb{N}$. Since $\ell_{n}\to s$ and $u_{n}\to s$ as $n\to\infty$, it follows from the squeeze theorem that $a_{n}\to s$ as $n\to\infty$.

(i) We argue by contradiction. Suppose $(a_{n})_{n\in\mathbb{N}}$ and $(b_{n})_{n\in\mathbb{N}}$ are convergent sequences with $a_{n}\leq b_{n}$ for all $n\in\mathbb{N}$ and

Let $\varepsilon:=(a-b)/2>0$. Then by the $\varepsilon$-$N$ definition of a limit there exists some $N_{1}\in\mathbb{N}$ such that $|a-a_{n}|<\varepsilon$ for all $n>N_{1}$. Similarly, there exists some $N_{2}\in\mathbb{N}$ such that $|b-b_{n}|<\varepsilon$ for all $n>N_{2}$. Taking $N:=\max\{N_{1},N_{2}\}$, it follows that $|a-a_{n}|<\varepsilon$ and $|b-b_{n}|<\varepsilon$ for all $n>N$. Consequently, for $n>N$ we have

and so

This is a contradiction, so we must have $a\leq b$.

(ii) This is false. Indeed, consider the convergent sequences $(a_{n})_{n\in\mathbb{N}}$ and $(b_{n})_{n\in\mathbb{N}}$ given by $a_{n}:=0$ and $b_{n}:=1/n$ for all $n\in\mathbb{N}$. Then $a_{n}<b_{n}$ for all $n\in\mathbb{N}$ but $\displaystyle\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}b_{n}=0$.

The proof of the boundedness test. If $a_{n}\to a$ as $n\to\infty$, then there exists some $N\in\mathbb{N}$ such that $|a_{n}|\leq|a|+1$ for all $n>N$. We can then take a maximum of the set of remaining terms to conclude that $|a_{n}|\leq M:=\max\{|a_{1}|,\dots,|a_{N-1}|,|a|+1\}$ for all $n\in\mathbb{N}$.

(i) For $n\in\mathbb{N}$, we have

Let $M>0$ be given and choose $N:=\lceil 2M\rceil$. If $n\in\mathbb{N}$ satisfies $n>N$, then

Thus, by definition, $\frac{n^{3}}{n+1}\to\infty$ as $n\to\infty$.

(ii) From Workshop 3, we know $n2^{-n}\to 0$ as $n\to\infty$. Hence, given $M>0$, taking $\varepsilon:=M^{-1}>0$ in the $\varepsilon$-$N$ definition of a limit, there exists some $N\in\mathbb{N}$ such that $0<n2^{-n}<M^{-1}$ for all $n>N$. Taking reciprocals, it follows that if $n>N$, then $2^{n}/n>M$. Thus, by definition, $2^{n}/n\to\infty$ as $n\to\infty$.

(iii) For $n\in\mathbb{N}$, by dividing through by $2^{2n}=4^{n}$, we have

By Exercise 2.31 and Worksheet 3, we know $(3/4)^{n}\to 0$ and $n^{2}4^{-n}\to 0$ as $n\to\infty$. By the limit laws, $(3/4)^{n}+n^{2}4^{-n}\to 0$ as $n\to\infty$. Hence, given $M>0$, taking $\varepsilon:=M^{-1}>0$ in the $\varepsilon$-$N$ definition of a limit, there exists some $N\in\mathbb{N}$ such that $0<(3/4)^{n}+n^{2}4^{-n}<M^{-1}$ for all $n>N$. Taking reciprocals, it follows that if $n>N$, then $\tfrac{2^{2n}}{3^{n}+n^{2}}>M$. Thus, by definition, $\tfrac{2^{2n}}{3^{n}+n^{2}}\to\infty$ as $n\to\infty$

Let $a_{n}:=(-1)^{n}n$ for all $n\in\mathbb{N}$. Then $(a_{n})_{n\in\mathbb{N}}$ is unbounded. Indeed, given $M\in\mathbb{R}$, if we choose $n>|M|$ odd, then it follows that $a_{n}=-n<-|M|\leq M$. On the other hand, if we chose $n>|M|$ even, then $a_{n}=n>|M|\geq M$. Thus, $M$ is neither a lower or an upper bound for $(a_{n})_{n\in\mathbb{N}}$. Since $M\in\mathbb{R}$ was chosen arbitrarily, it follows that $(a_{n})_{n\in\mathbb{N}}$ is not bounded above or below.

We claim that $a_{n}\not\to\infty$ as $n\to\infty$. Indeed, if $a_{n}\to\infty$ as $n\to\infty$, then there would exist some $N\in\mathbb{N}$ such that $a_{n}>0$ for all $n>N$. However, $a_{2n+1}=-(2n+1)<0$ for all $n\in\mathbb{N}$, so there can be no such choice of $N$.

Finally, we claim that $a_{n}\not\to-\infty$ as $n\to\infty$. Indeed, if $a_{n}\to-\infty$ as $n\to\infty$, then there would exist some $N\in\mathbb{N}$ such that $a_{n}<0$ for all $n>N$. However, $a_{2n}=2n>0$ for all $n\in\mathbb{N}$, so there can be no such choice of $N$.

Suppose $(a_{n})_{n\in\mathbb{N}}$ is sequence which converges to a limit $a\in\mathbb{R}$ and $(a_{n_{k}})_{k\in\mathbb{N}}$ is a subsequence. Let $\varepsilon>0$ be given. By applying the $\varepsilon$-$N$ definition of a limit to $(a_{n})_{n\in\mathbb{N}}$, there exists some $N\in\mathbb{N}$ such that $|a_{n}-a|<\varepsilon$ for all $n>N$. However, if $k>N$, then it follows that $n_{k}>k>N$ and so $|a_{n_{k}}-a|<\varepsilon$. Thus, by the $\varepsilon$-$N$ definition of a limit, $a_{n_{k}}\to a$ as $k\to\infty$.

Let $a_{n}:=\sin^{2}\big{(}\frac{\pi n}{50}\big{)}$ for $n\in\mathbb{N}$. Then

and

Hence, $a_{50k}\to 0$ as $k\to\infty$ and $a_{50k+25}\to 1$ as $k\to\infty$. In particular, we have found two subsequences with different limits, and so the sequence $(a_{n})_{n\in\mathbb{N}}$ must diverge by the subsequence test.

We show $(q_{n})_{n\in\mathbb{N}}$ diverges by the boundedness test. Indeed, let $M>0$ and choose $n>5M+5$. Then $q_{n}$ satisfies $n=5q_{n}+r_{n}$ where $0\leq r_{n}\leq 4$ and so

Thus, $M$ is not an upper bound for $(q_{n})_{n\in\mathbb{N}}$. Since $M>0$ was chosen arbitrarily, it follows that $(q_{n})_{n\in\mathbb{N}}$ is unbounded and therefore diverges by the boundedness test.

We show $(r_{n})_{n\in\mathbb{N}}$ diverges by the subsequence test. Indeed, for all $k\in\mathbb{N}$ we have $r_{5k}=0$ and $r_{5k+1}=1$. Thus, $r_{5k}\to 0$ and $r_{5k+1}\to 1$ as $k\to\infty$. Hence, $(r_{n})_{n\in\mathbb{N}}$ has two distinct subsequential limits and therefore diverges by the subsequence test.

Finally, we claim $\tfrac{r_{n}}{q_{n}+1}\to 0$ as $n\to\infty$. Indeed, let $\varepsilon>0$ and choose $N:=\lceil 20/\varepsilon\rceil+4$. If $n>N$, then arguing as in (A.3) we see that

Recalling that $0\leq r_{n}\leq 4$ for all $n\in\mathbb{N}$, we therefore have

Thus, by the $\varepsilon$-$N$ definition of a limit, $\tfrac{r_{n}}{q_{n}+1}\to 0$ as $n\to\infty$.

(i) Let $a\in\mathbb{R}$ and $\varepsilon:=1$. Given $N\in\mathbb{N}$, choose $n>\max\{N,|a|+1\}$, so that $n>N$. It follows that $|n-a|\geq|n|-|a|>1=\varepsilon$.

The above argument shows that for all $a\in\mathbb{R}$, there exists some $\varepsilon>0$ such that for all $N\in\mathbb{N}$ there exists some $n>N$ such that $|n-a|>\varepsilon$. Hence, by the $\varepsilon$-$N$ definition, $(n)_{n\in\mathbb{N}}$ diverges.

(ii) Let $a\in\mathbb{R}$ and $\varepsilon:=1/2$. Let $\sigma\in\{0,1\}$ satisfy

in other words, $\sigma:=0$ if $|1-a|\geq|1+a|$ and $\sigma:=1$ if $|1+a|\geq|1-a|$. It then follows from the definition that

Furthermore, by the triangle inequality,

and so $|(-1)^{\sigma}-a|\geq 1$.

Given $N\in\mathbb{N}$, choose $n>N$ even if $\sigma=0$ and $n>N$ odd if $\sigma=1$, so that $(-1)^{n}=(-1)^{\sigma}$. Thus,

The above argument shows that for all $a\in\mathbb{R}$, there exists some $\varepsilon>0$ such that for all $N\in\mathbb{N}$ there exists some $n>N$ such that $|(-1)^{n}-a|>\varepsilon$. Hence, by the $\varepsilon$-$N$ definition, $((-1)^{n})_{n\in\mathbb{N}}$ diverges.

Let $(a_{n})_{n\in\mathbb{N}}$ be non-increasing and bounded below. By Exercise 2.14 and Exercise 2.21, the sequence $(-a_{n})_{n\in\mathbb{N}}$ is non-decreasing and bounded above. Hence, by part 1 of the monotone convergence theorem, $(-a_{n})_{n\in\mathbb{N}}$ converges. Finally, by the limit laws, $(a_{n})_{n\in\mathbb{N}}$ = $(-(-a_{n}))_{n\in\mathbb{N}}$ converges.

(i) We prove $a_{n}>0$ for all $n\in\mathbb{N}$ by induction on $n$.

Base case: By hypothesis, $a_{1}=2>0$, which forms the base case of our induction.

Inductive step: Suppose, as an induction hypothesis, that $a_{n}>0$ for some $n\in\mathbb{N}$. Then using the formula

we see that $a_{n+1}$ is formed by multiplying together and summing positive terms and therefore must itself be positive. This closes the induction.

(ii) Let $n\in\mathbb{N}$. Using the formula

we expand out the square to give

as required. Since the squared term is always non-negative, we conclude that $a_{n+1}^{2}\geq 2$ for all $n\in\mathbb{N}$. Thus, since $a_{1}^{2}\geq 2$ by hypothesis, we have $a_{n}^{2}\geq 2$ for all $n\in\mathbb{N}$.

(iii) From part (i) we know the sequence $(a_{n})_{n\in\mathbb{N}}$ is bounded below by $0$. We claim $(a_{n})_{n\in\mathbb{N}}$ is monotone non-increasing. To see this, let $n\in\mathbb{N}$ and consider

By part (i), we know $a_{n}>0$ and by part (ii) we know $a_{n}^{2}-2\geq 0$. Hence the right-hand side of (A.4) is non-negative and so $a_{n}\geq a_{n+1}$. Thus the sequence is monotone non-increasing.