3.11.E: Problems on Limits of Sequences (Exercises)

See also Chapter 2, §13.

Exercise $\PageIndex{1}$

Prove that if $x_{m} \rightarrow 0$ and if $\left\{a_{m}\right\}$ is bounded in $E^{1}$ or $C,$ then
$$a_{m} x_{m} \rightarrow 0.$$
This is true also if the $x_{m}$ are vectors and the $a_{m}$ are scalars (or vice versa).
[Hint: If $\left\{a_{m}\right\}$ is bounded, there is a $K \in E^{1}$ such that
$$(\forall m) \quad\left|a_{m}\right|<K.$$
As $x_{m} \rightarrow 0$,
$$(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad\left|x_{m}\right|<\frac{\varepsilon}{K}(\mathrm{why} ?),$$
so $\left|a_{m} x_{m}\right|<\varepsilon . ]$

Exercise $\PageIndex{2}$

Prove Theorem 1$(\text { ii })$.
[Hint: By Corollary 2(ii)(iii) in §14, we must show that $a_{m} x_{m}-a w \rightarrow 0$. Now
$$a_{m} x_{m}-a q=a_{m}\left(x_{m}-q\right)+\left(a_{m}-a\right) q.$$
where $x_{m}-q \rightarrow 0$ and $a_{m}-a \rightarrow 0$ by Corollary 2 of §14. Hence by Problem 1,
$$a_{m}\left(x_{m}-q\right) \rightarrow 0 \text { and }\left(a_{m}-a\right) q \rightarrow 0$$
(treat $q$ as a constant sequence and use Corollary 5 in §14). Now apply Theorem 1$(\mathrm{i}) . ]$

Exercise $\PageIndex{3}$

Prove that if $a_{m} \rightarrow a$ and $a \neq 0$ in $E^{1}$ or $C,$ then
$$(\exists \varepsilon>0)(\exists k)(\forall m>k) \quad\left|a_{m}\right| \geq \varepsilon.$$
(We briefly say that the $a_{m}$ are bounded away from $0,$ for $m>k . )$ Hence prove the boundedness of $\left\{\frac{1}{a_{m}}\right\}$ for $m>k$.
[Hint: For the first part, proceed as in the proof of Corollary 1 in $§14, \text { with } x_{m}=a_{m},$ $p=a,$ and $q=0 .$
For the second part, the inequalities
$$(\forall m>k) \quad\left|\frac{1}{a_{m}}\right| \leq \frac{1}{\varepsilon}$$
lead to the desired result. $]$

Exercise $\PageIndex{4}$

Prove that if $a_{m} \rightarrow a \neq 0$ in $E^{1}$ or $C,$ then
$$\frac{1}{a_{m}} \rightarrow \frac{1}{a}.$$
Use this and Theorem 1$(\text { ii) to prove Theorem } 1(\text { iii), noting that }$
$$\frac{x_{m}}{a_{m}}=x_{m} \cdot \frac{1}{a_{m}}.$$
[Hint: Use Note 3 and Problem 3 to find that
$$(\forall m>k) \quad\left|\frac{1}{a_{m}}-\frac{1}{a}\right|=\frac{1}{|a|}\left|a_{m}-a\right| \frac{1}{\left|a_{m}\right|},$$
where $\left\{\frac{1}{a_{m}}\right\}$ is bounded and $\frac{1}{|a|}\left|a_{m}-a\right| \rightarrow 0 .$ (Why?)
Hence, by Problem $1,\left|\frac{1}{a_{m}}-\frac{1}{a}\right| \rightarrow 0 .$ Proceed. $]$

Exercise $\PageIndex{5}$

Prove Corollaries 1 and 2 in two ways:
(i) Use Definition 2 of Chapter 2, §13 for Corollary $1(a),$ treating infinite limits separately; then prove (b) by assuming the opposite and exhibiting a contradiction to $(a) .$
(ii) Prove (b) first by using Corollary 2 and Theorem 3 of Chapter 2, §13; then deduce (a) by contradiction.

Exercise $\PageIndex{6}$

Prove Corollary 3 in two ways (cf. Problem 5).

Exercise $\PageIndex{7}$

Prove Theorem 4 as suggested, and also without using Theorem 1$(\mathrm{i})$.

Exercise $\PageIndex{8}$

Prove Theorem 2.
[Hint: If $\overline{x}_{m} \rightarrow \overline{p},$ then
$$(\forall \varepsilon>0)(\exists q)(\forall m>q) \quad \varepsilon>\left|\overline{x}_{m}-\overline{p}\right| \geq\left|x_{m k}-p_{k}\right| . \quad(\mathrm{Why} ?)$$
Thus by definition $x_{m k} \rightarrow p_{k}, k=1,2, \ldots, n$.
Conversely, if so, use Theorem 1$(\mathrm{i})(\text { ii })$ to obtain
$$\sum_{k=1}^{n} x_{m k} \vec{e}_{k} \rightarrow \sum_{k=1}^{n} p_{k} \vec{e}_{k},$$
with $\vec{e}_{k}$ as in Theorem 2 of §§1-3].

Exercise $\PageIndex{8'}$

In Problem $8,$ prove the converse part from definitions. $(\text { Fix } \varepsilon>0, \text { etc. })$

Exercise $\PageIndex{9}$

Find the following limits in $E^{1},$ in two ways: (i) using Theorem 1, justifying each step; (ii) using definitions only.
$$\begin{array}{ll}{\text { (a) } \lim _{m \rightarrow \infty} \frac{m+1}{m} ;} & {\text { (b) } \lim _{m \rightarrow \infty} \frac{3 m+2}{2 m-1}} \\ {\text { (c) } \lim _{n \rightarrow \infty} \frac{1}{1+n^{2}} ;} & {\text { (d) } \lim _{n \rightarrow \infty} \frac{n(n-1)}{1-2 n^{2}}}\end{array}$$
$[\text { Solution of }(\mathrm{a}) \text { by the first method: Treat }$
$$\frac{m+1}{m}=1+\frac{1}{m}$$
as the sum of $x_{m}=1$ (constant) and
$$y_{m}=\frac{1}{m} \rightarrow 0 \text { (proved in } § 14 ).$$
Thus by Theorem 1$(\mathrm{i})$,
$$\frac{m+1}{m}=x_{m}+y_{m} \rightarrow 1+0=1.$$
Second method: Fix $\varepsilon>0$ and find $k$ such that
$$(\forall m>k) \quad\left|\frac{m+1}{m}-1\right|<\varepsilon .$$
Solving for $m,$ show that this holds if $m>\frac{1}{\varepsilon} .$ Thus take an integer $k>\frac{1}{\varepsilon},$ so
$$(\forall m>k) \quad\left|\frac{m+1}{m}-1\right|<\varepsilon.$$
Caution: One cannot apply Theorem 1 (iii) directly, treating $(m+1) / m$ as the quotient of $x_{m}=m+1$ and $a_{m}=m,$ because $x_{m}$ and $a_{m}$ diverge in $E^{1} .$ (Theorem 1 does not apply to infinite limits.) As a remedy, we first divide the numerator and denominator by a suitable power of $m(\text { or } n) . ]$

Exercise $\PageIndex{10}$

Prove that
$$\left|x_{m}\right| \rightarrow+\infty \text { in } E^{*} \text { iff } \frac{1}{x_{m}} \rightarrow 0 \quad\left(x_{m} \neq 0\right).$$

Exercise $\PageIndex{11}$

Prove that if
$$x_{m} \rightarrow+\infty \text { and } y_{m} \rightarrow q \neq-\infty \text { in } E^{*},$$
then
$$x_{m}+y_{m} \rightarrow+\infty.$$
This is written symbolically as
$$" +\infty+q=+\infty \text { if } q \neq-\infty ."$$
Do also
$$" -\infty+q=-\infty \text { if } q \neq+\infty . "$$
Prove similarly that
$$"(+\infty) \cdot q=+\infty \text { if } q>0"$$
and
$$"(+\infty) \cdot q=-\infty \text { if } q<0."$$
[Hint: Treat the cases $q \in E^{1}, q=+\infty,$ and $q=-\infty$ separately. Use definitions.]

Exercise $\PageIndex{12}$

Find the limit (or $\underline{\lim}$ and $\overline{\lim}$) of the following sequences in $E^{*} :$
(a) $x_{n}=2 \cdot 4 \cdots 2 n=2^{n} n !$;
(b) $x_{n}=5 n-n^{3} ;$
(c) $x_{n}=2 n^{4}-n^{3}-3 n^{2}-1$;
(d) $x_{n}=(-1)^{n} n !$;
(e) $x_{n}=\frac{(-1)^{n}}{n !}$.
[Hint for $(\mathrm{b}) : x_{n}=n\left(5-n^{2}\right) ;$ use Problem 11.]

Exercise $\PageIndex{13}$

Use Corollary 4 in §14, to find the following:
(a) $\lim _{n \rightarrow \infty} \frac{(-1)^{n}}{1+n^{2}}$;
(b) $\lim _{n \rightarrow \infty} \frac{1-n+(-1)^{n}}{2 n+1}$.

Exercise $\PageIndex{14}$

Find the following.
(a) $\lim _{n \rightarrow \infty} \frac{1+2+\cdots+n}{n^{2}}$;
(b) $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{2}}{n^{3}+1}$;
(c) $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{3}}{n^{4}-1}$.
[Hint: Compute $\sum_{k=1}^{n} k^{m}$ using Problem 10 of Chapter 2, §§5-6.]
What is wrong with the following "solution" of $(a) : \frac{1}{n^{2}} \rightarrow 0, \frac{2}{n^{2}} \rightarrow 0,$ etc.; hence the limit is 0$?$

Exercise $\PageIndex{15}$

For each integer $m \geq 0,$ let
$$S_{m n}=1^{m}+2^{m}+\cdots+n^{m}.$$
Prove by induction on $m$ that
$$\lim _{n \rightarrow \infty} \frac{S_{m n}}{(n+1)^{m+1}}=\frac{1}{m+1}.$$
[Hint: First prove that
$$(m+1) S_{m n}=(n+1)^{m+1}-1-\sum_{i=0}^{m-1}\left(\begin{array}{c}{m+1} \\ {i}\end{array}\right) S_{m i}$$
by adding up the binomial expansions of $(k+1)^{m+1}, k=1, \ldots, n . ]$

Exercise $\PageIndex{16}$

Prove that
$$\lim _{n \rightarrow \infty} q^{n}=+\infty \text { if } q>1 ; \quad \lim _{n \rightarrow \infty} q^{n}=0 \text { if }|q|<1 ; \quad \lim _{n \rightarrow \infty} 1^{n}=1.$$
[Hint: If $q>1,$ put $q=1+d, d>0 .$ By the binomial expansion,
$$q^{n}=(1+d)^{n}=1+n d+\cdots+d^{n}>n d \rightarrow+\infty . \quad(\mathrm{Why?})$$
If $|q|<1,$ then $\left|\frac{1}{q}\right|>1 ;$ so $\lim \left|\frac{1}{q}\right|^{n}=+\infty ;$ use Problem $10 . ]$

Exercise $\PageIndex{17}$

Prove that
$$\lim _{n \rightarrow \infty} \frac{n}{q^{n}}=0 \text { if }|q|>1, \text { and } \lim _{n \rightarrow \infty} \frac{n}{q^{n}}=+\infty \text { if } 0<q<1.$$
[Hint: If $|q|>1,$ use the binomial as in Problem 16 to obtain
$$|q|^{n}>\frac{1}{2} n(n-1) d^{2}, n \geq 2, \text { so } \frac{n}{|q|^{n}}<\frac{2}{(n-1) d^{2}} \rightarrow 0.$$
Use Corollary 3 with
$$x_{n}=0,\left|z_{n}\right|=\frac{n}{|q|^{n}}, \text { and } y_{n}=\frac{2}{(n-1) d^{2}}$$
to get $\left|z_{n}\right| \rightarrow 0 ;$ hence also $z_{n} \rightarrow 0$ by Corollary 2$(\text { iii) of } §14 . \text { In case } 0<q<1, \text { use }$ 10.]

Exercise $\PageIndex{18}$

Let $r, a \in E^{1} .$ Prove that
$$\lim _{n \rightarrow \infty} n^{r} a^{-n}=0 \text { if }|a|>1.$$
[Hint: If $r>1$ and $a>1,$ use Problem 17 with $q=a^{1 / r}$ to get $n a^{-n / r} \rightarrow 0 .$ As
$$0<n^{r} a^{-n}=\left(n a^{-n / r}\right)^{r} \leq n a^{-n / r} \rightarrow 0,$$
obtain $n^{r} a^{-n} \rightarrow 0$.
If $r<1,$ then $n^{r} a^{-n}<n a^{-n} \rightarrow 0 .$ What if $a<-1 ? ]$

Exercise $\PageIndex{19}$

(Geometric series.) Prove that if $|q|<1,$ then
$$\lim _{n \rightarrow \infty}\left(a+a q+\cdots+a q^{n-1}\right)=\frac{a}{1-q}.$$
[Hint:
$$a\left(1+q+\cdots+q^{n-1}\right)=a \frac{1-q^{n}}{1-q},$$
where $q^{n} \rightarrow 0,$ by Problem $16 . ]$

Exercise $\PageIndex{20}$

Let $0<c<+\infty .$ Prove that
$$\lim _{n \rightarrow \infty} \sqrt[n]{c}=1.$$
$\left[\text { Hint: If } c>1, \text { put } \sqrt[n]{c}=1+d_{n}, d_{n}>0 . \text { Expand } c=\left(1+d_{n}\right)^{n} \text { to show that }\right.$
$$0<d_{n}<\frac{c}{n} \rightarrow 0,$$
so $d_{n} \rightarrow 0$ by Corollary $3 . ]$

Exercise $\PageIndex{21}$

Investigate the following sequences for monotonicity, $\underline{\lim}$, $\overline{\lim}$, and $\lim$. (In each case, find suitable formula, or formulas, for the general term.)
(a) $2,5,10,17,26, \ldots$;
(b) $2,-2,2,-2, \ldots$;
(c) $2,-2,-6,-10,-14, \ldots ;$
(d) $1,1,-1,-1,1,1,-1,-1, \ldots ;$
(e) $\frac{3 \cdot 2}{1}, \frac{4 \cdot 6}{4}, \frac{5 \cdot 10}{9}, \frac{6 \cdot 14}{16}, \ldots$.

Exercise $\PageIndex{22}$

Do Problem 21 for the following sequences.
(a) $\frac{1}{2 \cdot 3}, \frac{-8}{3 \cdot 4}, \frac{27}{4 \cdot 5}, \frac{-64}{5 \cdot 6}, \frac{125}{6 \cdot 7}, \ldots ;$
(b) $\frac{2}{9},-\frac{5}{9}, \frac{8}{9},-\frac{13}{9}, \ldots ;$
(c) $\frac{2}{3},-\frac{2}{5}, \frac{4}{7},-\frac{4}{9}, \frac{6}{11},-\frac{6}{13}, \ldots$
(d) $1,3,5,1,1,3,5,2,1,3,5,3, \ldots, 1,3,5, n, \ldots ;$
(e) $0.9,0.99,0.999, \ldots$;
(f) $+\infty, 1,+\infty, 2,+\infty, 3, \dots ;$
$(\mathrm{g})-\infty, 1,-\infty, \frac{1}{2}, \ldots,-\infty, \frac{1}{n}, \ldots$.

Exercise $\PageIndex{23}$

Do Problem 20 as follows: If $c \geq 1,\{\sqrt[n]{c}\} \downarrow .(\mathrm{Why} ?)$ By Theorem $3,$ $p=\lim _{n \rightarrow \infty} \sqrt[n]{c}$ exists and
$$(\forall n) \quad 1 \leq p \leq \sqrt[n]{c}, \text { i.e., } 1 \leq p^{n} \leq c .$$
By Problem $16, p$ cannot be $>1,$ so $p=1$.
In case $0<c<1,$ consider $\sqrt[n]{1 / c}$ and use Theorem 1$(\text { iii) }$.

Exercise $\PageIndex{24}$

Prove the existence of $\lim x_{n}$ and find it when $x_{n}$ is defined inductively by
(i) $x_{1}=\sqrt{2}, x_{n+1}=\sqrt{2 x_{n}}$;
(ii) $x_{1}=c>0, x_{n+1}=\sqrt{c^{2}+x_{n}}$;
(iii) $x_{1}=c>0, x_{n+1}=\frac{c x_{n}}{n+1} ;$ hence deduce that $\lim _{n \rightarrow \infty} \frac{c^{n}}{n !}=0$.
[Hint: Show that the sequences are monotone and bounded in $E^{1}$ (Theorem 3).
For example, in (ii) induction yields
$$x_{n}<x_{n+1}<c+1 . \quad(\text { Verify! })$$
Thus $\lim x_{n}=\lim x_{n+1}=p$ exists. To find $p,$ square the equation
$$x_{n+1}=\sqrt{c^{2}+x_{n}} \quad(\text { given })$$
and use Theorem 1 to get
$$p^{2}=c^{2}+p . \quad(\mathrm{Why?})$$
Solving for $p$ (noting that $p>0 ),$ obtain
$$p=\lim x_{n}=\frac{1}{2}\left(1+\sqrt{4 c^{2}+1}\right);$$
similarly in cases (i) and (iii). $]$

Exercise $\PageIndex{25}$

Find $\lim x_{n}$ in $E^{1}$ or $E^{*}$ (if any), given that
(a) $x_{n}=(n+1)^{q}-n^{q}, 0<q<1$;
(b) $x_{n}=\sqrt{n}(\sqrt{n+1}-\sqrt{n})$;
(c) $x_{n}=\frac{1}{\sqrt{n^{2}+k}}$;
(d) $x_{n}=n(n+1) c^{n},$ with $|c|<1$;
(e) $x_{n}=\sqrt[n]{\sum_{k=1}^{m} a_{k}^{n}},$ with $a_{k}>0$;
(f) $x_{n}=\frac{3 \cdot 5 \cdot 7 \cdots(2 n+1)}{2 \cdot 5 \cdot 8 \cdots(3 n-1)}$.
[Hints:
(a) $0<x_{n}=n^{q}\left[\left(1+\frac{1}{n}\right)^{q}-1\right]<n^{q}\left(1+\frac{1}{n}-1\right)=n^{q-1} \rightarrow 0 .(\mathrm{Why} ?)$
(b) $x_{n}=\frac{1}{1+\sqrt{1+1 / n}},$ where $1<\sqrt{1+\frac{1}{n}}<1+\frac{1}{n} \rightarrow 1,$ so $x_{n} \rightarrow \frac{1}{2} .$ (Why?)
(c) Verify that
$$\frac{n}{\sqrt{n^{2}+n}} \leq x_{n} \leq \frac{n}{\sqrt{n^{2}+1}},$$
so $x_{n} \rightarrow 1$ by Corollary 3. (Give a proof.)
(d) See Problems 17 and 18.
(e) Let $a=\max \left(a_{1}, \ldots, a_{m}\right) .$ Prove that $a \leq x_{n} \leq a \sqrt[n]{m} .$ Use Problem $20 . ]$
The following are some harder but useful problems of theoretical importance.
The explicit hints should make them not too hard.

Exercise $\PageIndex{26}$

Let $\left\{x_{n}\right\} \subseteq E^{1} .$ Prove that if $x_{n} \rightarrow p$ in $E^{1},$ then also
$$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} x_{i}=p$$
(i.e., $p$ is also the limit of the sequence of the arithmetic means of the $x_{n} ).$
[Solution: Fix $\varepsilon>0 .$ Then
$$(\exists k)(\forall n>k) \quad p-\frac{\varepsilon}{4}<x_{n}<p+\frac{\varepsilon}{4}.$$
Adding $n-k$ inequalities, get
$$(n-k)\left(p-\frac{\varepsilon}{4}\right)<\sum_{i=k+1}^{n} x_{i}<(n-k)\left(p+\frac{\varepsilon}{4}\right).$$
With $k$ so fixed, we thus have
$$(\forall n>k) \quad \frac{n-k}{n}\left(p-\frac{\varepsilon}{4}\right)<\frac{1}{n}\left(x_{k+1}+\cdots+x_{n}\right)<\frac{n-k}{n}\left(p+\frac{\varepsilon}{4}\right).$$
Here, with $k$ and $\varepsilon$ fixed,
$$\lim _{n \rightarrow \infty} \frac{n-k}{n}\left(p-\frac{\varepsilon}{4}\right)=p-\frac{\varepsilon}{4}.$$
Hence, as $p-\frac{1}{2} \varepsilon<p-\frac{1}{4} \varepsilon,$ there is $k^{\prime}$ such that
$$\left(\forall n>k^{\prime}\right) \quad p-\frac{\varepsilon}{2}<\frac{n-k}{n}\left(p-\frac{\varepsilon}{4}\right).$$
Similarly,
$$\left(\exists k^{\prime \prime}\right)\left(\forall n>k^{\prime \prime}\right) \quad \frac{n-k}{n}\left(p+\frac{\varepsilon}{4}\right)<p+\frac{\varepsilon}{2}.$$
Combining this with (i), we have, for $K^{\prime}=\max \left(k, k^{\prime}, k^{\prime \prime}\right)$,
$$\left(\forall n>K^{\prime}\right) \quad p-\frac{\varepsilon}{2}<\frac{1}{n}\left(x_{k+1}+\cdots+x_{n}\right)<p+\frac{\varepsilon}{2}.$$
Now with $k$ fixed,
$$\lim _{n \rightarrow \infty} \frac{1}{n}\left(x_{1}+x_{2}+\cdots+x_{k}\right)=0.$$
Hence
$$\left(\exists K^{\prime \prime}\right)\left(\forall n>K^{\prime \prime}\right) \quad-\frac{\varepsilon}{2}<\frac{1}{n}\left(x_{1}+\cdots+x_{k}\right)<\frac{\varepsilon}{2}.$$
Let $K=\max \left(K^{\prime}, K^{\prime \prime}\right) .$ Then combining with (ii), we have
$$(\forall n>K) \quad p-\varepsilon<\frac{1}{n}\left(x_{1}+\cdots+x_{n}\right)<p+\varepsilon,$$
and the result follows.

Exercise $\PageIndex{26'}$

Show that the result of Problem 26 holds also for infinite limits $p=\pm \infty \in E^{*} .$

Exercise $\PageIndex{27}$

Prove that if $x_{n} \rightarrow p$ in $E^{*}\left(x_{n}>0\right),$ then
$$\lim _{n \rightarrow \infty} \sqrt[n]{x_{1} x_{2} \cdots x_{n}}=p.$$
[Hint: Let first $0<p<+\infty .$ Given $\varepsilon>0,$ use density to fix $\delta>1$ so close to 1 that
$$p-\varepsilon<\frac{p}{\delta}<p<p \delta<p+\varepsilon.$$
As $x_{n} \rightarrow p$,
$$(\exists k)(\forall n>k) \quad \frac{p}{\sqrt[4]{\delta}}<x_{n}<p \sqrt[4] \delta.$$
Continue as in Problem $26,$ replacing $\varepsilon$ by $\delta,$ and multiplication by addition (also subtraction by division, etc., as shown above). Find a similar solution for the case $p=+\infty .$ Note the result of Problem 20.]

Exercise $\PageIndex{28}$

Disprove by counterexamples the converse implications in Problems 26 and $27 .$ For example, consider the sequences
$$1,-1,1,-1, \dots$$
and
$$\frac{1}{2}, 2, \frac{1}{2}, 2, \frac{1}{2}, 2, \ldots$$

Exercise $\PageIndex{29}$

Prove the following.
(i) If $\left\{x_{n}\right\} \subset E^{1}$ and $\lim _{n \rightarrow \infty}\left(x_{n+1}-x_{n}\right)=p$ in $E^{*},$ then $\frac{x_{n}}{n} \rightarrow p$.
(ii) If $\left\{x_{n}\right\} \subset E^{1}\left(x_{n}>0\right)$ and if $\frac{x_{n+1}}{x_{n}} \rightarrow p \in E^{*},$ then $\sqrt[n]{x_{n}} \rightarrow p$.
Disprove the converse statements by counterexamples.
[Hint: For $(\mathrm{i}),$ let $y_{1}=x_{1}$ and $y_{n}=x_{n}-x_{n-1}, n=2,3, \ldots$ Then $y_{n} \rightarrow p$ and
$$\frac{1}{n} \sum_{i=1}^{n} y_{i}=\frac{x_{n}}{n},$$
so Problems 26 and $26^{\prime}$ apply.
For (ii), use Problem $27 .$ See Problem 28 for examples. $]$

Exercise $\PageIndex{30}$

From Problem 29 deduce that
(a) $\lim _{n \rightarrow \infty} \sqrt[n]{n !}=+\infty$;
(b) $\lim _{n \rightarrow \infty} \frac{n+1}{n !}=0$;
(c) $\lim _{n \rightarrow \infty} \sqrt[n]{\frac{n^{n}}{n !}}=e$;
(d) $\lim _{n \rightarrow \infty} \frac{1}{n} \sqrt[n]{n !}=\frac{1}{e}$;
(e) $\lim _{n \rightarrow \infty} \sqrt[n]{n}=1$.

Exercise $\PageIndex{31}$

Prove that
$$\lim _{n \rightarrow \infty} x_{n}=\frac{a+2 b}{3},$$
given
$$x_{0}=a, x_{1}=b, \text { and } x_{n+2}=\frac{1}{2}\left(x_{n}+x_{n+1}\right).$$
[Hint: Show that the differences $d n=x_{n}-x_{n-1}$ form a geometric sequence, with ratio $q=-\frac{1}{2},$ and $x_{n}=a+\sum_{k=1}^{n} d_{k} .$ Then use the result of Problem $19 . ]$

Exercise $\PageIndex{32}$

$\Rightarrow 32 .$ For any sequence $\left\{x_{n}\right\} \subseteq E^{1},$ prove that
$$\underline{\lim} x_{n} \leq \underline{\lim} \frac{1}{n} \sum_{i = 1}^{n} x_{i} \leq \overline{\lim} \frac{1}{n} \sum_{i = 1}^{n} x_{i} \leq \overline{\lim} x_{n} .$$
Hence find a new solution of Problems 26 and $26^{\prime} .$
[Proof for $\overline{\lim}$: Fix any $k \in N .$ Put
$$c=\sum_{i=1}^{k} x_{i} \text { and } b=\sup _{i \geq k} x_{i}.$$
Verify that
$$(\forall n>k) \quad x_{k+1}+x_{k+2}+\cdots+x_{n} \leq(n-k) b.$$
Add $c$ on both sides and divide by $n$ to get
$$(\forall n>k) \quad \frac{1}{n} \sum_{i=1}^{n} x_{i} \leq \frac{c}{n}+\frac{n-k}{n} b.$$
Now fix any $\varepsilon>0,$ and first let $|b|<+\infty .$ As $\frac{c}{n} \rightarrow 0$ and $\frac{n-k}{n} b \rightarrow b,$ there is $n_{k}>k$ such that
$$\left(\forall n>n_{k}\right) \quad \frac{c}{n}<\frac{\varepsilon}{2} \text { and } \frac{n-k}{n} b<b+\frac{\varepsilon}{2}.$$
Thus by $\left(\mathrm{i}^{*}\right)$,
$$\left(\forall n>n_{k}\right) \quad \frac{1}{n} \sum_{i=1}^{n} x_{i} \leq \varepsilon+b.$$
This clearly holds also if $b=\sup _{i \geq k} x_{i}=+\infty .$ Hence also
$$\sup _{n \geq n_{k}} \frac{1}{n} \sum_{i=1}^{n} x_{i} \leq \varepsilon+\sup _{i \geq k} x_{i}.$$
As $k$ and $\varepsilon$ were arbitrary, we may let first $k \rightarrow+\infty,$ then $\varepsilon \rightarrow 0,$ to obtain
$$\underline{\lim} \frac{1}{n} \sum_{i=1}^{n} x_{i} \leq \lim _{k \rightarrow \infty} \sup _{i \geq k} x_{i}=\overline{\lim } x_{n} . \quad(\text { Explain! }) ]$$

Exercise $\PageIndex{33}$

$\Rightarrow 33 .$ Given $\left\{x_{n}\right\} \subseteq E^{1}, x_{n}>0,$ prove that
$$\underline{\lim} x_{n} \leq \underline{\lim} \sqrt[n]{x_{1} x_{2} \cdots x_{n}} \text{ and } \overline{\lim} \sqrt[n]{x_{1} x_{2} \cdots x_{n}} \leq \overline{\lim} x_{n} .$$
Hence obtain a new solution for Problem $27 .$
[Hint: Proceed as suggested in Problem $32,$ replacing addition by multiplication.]

Exercise $\PageIndex{34}$

Given $x_{n}, y_{n} \in E^{1}\left(y_{n}>0\right),$ with
$$x_{n} \rightarrow p \in E^{*} \text { and } b_{n}=\sum_{i=1}^{n} y_{i} \rightarrow+\infty,$$
prove that
$$\lim _{n \rightarrow \infty} \frac{\sum_{i=1}^{n} x_{i} y_{i}}{\sum_{i=1}^{n} y_{i}}=p.$$
Note that Problem 26 is a special case of Problem 34 (take all $y_{n}=1 )$. [Hint for a finite $p :$ Proceed as in Problem $26 .$ However, before adding the $n-k$ inequalities, multiply by $y_{i}$ and obtain
$$\left(p-\frac{\varepsilon}{4}\right) \sum_{i=k+1}^{n} y_{i}<\sum_{i=k+1}^{n} x_{i} y_{i}<\left(p+\frac{\varepsilon}{4}\right) \sum_{i=k+1}^{n} y_{i}.$$
$\operatorname{Put} b_{n}=\sum_{i=1}^{n} y_{i}$ and show that
$$\frac{1}{b_{n}} \sum_{i=k+1}^{n} x_{i} y_{i}=1-\frac{1}{b_{n}} \sum_{i=1}^{k} x_{i} y_{i},$$
where $b_{n} \rightarrow+\infty(\text { by assumption }),$ so
$$\frac{1}{b_{n}} \sum_{i=1}^{k} x_{i} y_{i} \rightarrow 0 \quad \text { (for a fixed } k ).$$
Proceed. Find a proof for $p=\pm \infty . ]$

Exercise $\PageIndex{35}$

Do Problem 34 by considering $\underline{\lim}$ and $\overline{\lim}$ as in Problem 32.
$\left[\text { Hint: Replace } \frac{c}{n} \text { by } \frac{c}{b_{n}}, \text { where } b_{n}=\sum_{i=1}^{n} y_{i} \rightarrow+\infty .\right]$

Exercise $\PageIndex{36}$

Prove that if $u_{n}, v_{n} \in E^{1},$ with $\left\{v_{n}\right\} \uparrow$ (strictly) and $v_{n} \rightarrow+\infty,$ and if
$$\lim _{n \rightarrow \infty} \frac{u_{n}-u_{n-1}}{v_{n}-v_{n-1}}=p \quad\left(p \in E^{*}\right),$$
then also
$$\lim _{n \rightarrow \infty} \frac{u_{n}}{v_{n}}=p,$$
[Hint: The result of Problem $34,$ with
$$x_{n}=\frac{u_{n}-u_{n-1}}{v_{n}-v_{n-1}} \text { and } y_{n}=v_{n}-v_{n-1}.$$
leads to the final result. $]$

Exercise $\PageIndex{37}$

From Problem 36 obtain a new solution for Problem $15 .$ Also prove that
$$\lim _{n \rightarrow \infty}\left(\frac{S_{m n}}{n^{m+1}}-\frac{1}{m+1}\right)=\frac{1}{2}.$$
[Hint: For the first part, put
$$u_{n}=S_{m n} \text { and } v_{n}=n^{m+1}.$$
For the second, put
$$u_{n}=(m+1) S_{m n}-n^{m+1} \text { and } v_{n}=n^{m}(m+1) . ]$$

Exercise $\PageIndex{38}$

Let $0<a<b<+\infty .$ Define inductively: $a_{1}=\sqrt{a b}$ and $b_{1}=\frac{1}{2}(a+b)$;
$$a_{n+1}=\sqrt{a_{n} b_{n}} \text { and } b_{n+1}=\frac{1}{2}\left(a_{n}+b_{n}\right), n=1,2, \ldots$$
Then $a_{n+1}<b_{n+1}$ for
$$b_{n+1}-a_{n+1}=\frac{1}{2}\left(a_{n}+b_{n}\right)-\sqrt{a_{n} b_{n}}=\frac{1}{2}\left(\sqrt{b_{n}}-\sqrt{a_{n}}\right)^{2}>0.$$
Deduce that
$$a<a_{n}<a_{n+1}<b_{n+1}<b_{n}<b,$$
so $\left\{a_{n}\right\} \uparrow$ and $\left\{b_{n}\right\} \downarrow .$ By Theorem $3, a_{n} \rightarrow p$ and $b_{n} \rightarrow q$ for some $p, q \in E^{1} .$ Prove that $p=q,$ i.e.,
$$\lim a_{n}=\lim b_{n}.$$
(This is Gauss's arithmetic-geometric mean of $a$ and $b . )$
[Hint: Take limits of both sides in $b_{n+1}=\frac{1}{2}\left(a_{n}+b_{n}\right)$ to get $q=\frac{1}{2}(p+q) . ]$

Exercise $\PageIndex{39}$

Let $0<a<b$ in $E^{1} .$ Define inductively $a_{1}=a, b_{1}=b$,
$$a_{n+1}=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}, \text { and } b_{n+1}=\frac{1}{2}\left(a_{n}+b_{n}\right), \quad n=1,2, \ldots$$
Prove that
$$\sqrt{a b}=\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}.$$
[Hint: Proceed as in Problem 38.]

Exercise $\PageIndex{40}$

Prove the continuity of dot multiplication, namely, if
$$\overline{x}_{n} \rightarrow \overline{q} \text { and } \overline{y}_{n} \rightarrow \overline{r} \text { in } E^{n}$$
(*or in another Euclidean space; see §9), then
$$\overline{x}_{n} \cdot \overline{y}_{n} \rightarrow \overline{q} \cdot \overline{r}.$$