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


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
$\overline{x}_{n} \cdot \overline{y}_{n} \rightarrow \overline{q} \cdot \overline{r}.$