3.11.E: Problems on Limits of Sequences (Exercises)
See also Chapter 2, §13.
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 \(\left|a_{m} x_{m}\right|<\varepsilon . ]\)
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\((\mathrm{i}) . ]\)
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. \(]\)
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. \(]\)
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.
Prove Corollary 3 in two ways (cf. Problem 5).
Prove Theorem 4 as suggested, and also without using Theorem 1\((\mathrm{i})\).
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].
In Problem \(8,\) prove the converse part from definitions. \((\text { Fix } \varepsilon>0, \text { etc. })\)
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 \(m(\text { or } n) . ]\)
Prove that
\[
\left|x_{m}\right| \rightarrow+\infty \text { in } E^{*} \text { iff } \frac{1}{x_{m}} \rightarrow 0 \quad\left(x_{m} \neq 0\right).
\]
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.]
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.]
Use Corollary 4 in §14, to find the following:
(a) \(\lim _{n \rightarrow \infty} \frac{(-1)^{n}}{1+n^{2}}\);
(b) \(\lim _{n \rightarrow \infty} \frac{1-n+(-1)^{n}}{2 n+1}\).
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\(?\)
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 \((k+1)^{m+1}, k=1, \ldots, n . ]\)
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 \(10 . ]\)
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.]
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 \(a<-1 ? ]\)
(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 \(16 . ]\)
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 \(3 . ]\)
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) \(\frac{3 \cdot 2}{1}, \frac{4 \cdot 6}{4}, \frac{5 \cdot 10}{9}, \frac{6 \cdot 14}{16}, \ldots\).
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) \(+\infty, 1,+\infty, 2,+\infty, 3, \dots ;\)
\((\mathrm{g})-\infty, 1,-\infty, \frac{1}{2}, \ldots,-\infty, \frac{1}{n}, \ldots\).
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\((\text { iii) }\).
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). \(]\)
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.
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.
Show that the result of Problem 26 holds also for infinite limits \(p=\pm \infty \in E^{*} .\)
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.]
Disprove by counterexamples the converse implications in Problems 26 and \(27 .\) For example, consider the sequences
\[
1,-1,1,-1, \dots
\]
and
\[
\frac{1}{2}, 2, \frac{1}{2}, 2, \frac{1}{2}, 2, \ldots
\]
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. \(]\)
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) \(\lim _{n \rightarrow \infty} \sqrt[n]{n}=1\).
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 \(19 . ]\)
\(\Rightarrow 32 .\) 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
\[
\underline{\lim} \frac{1}{n} \sum_{i=1}^{n} x_{i} \leq \lim _{k \rightarrow \infty} \sup _{i \geq k} x_{i}=\overline{\lim } x_{n} . \quad(\text { Explain! }) ]
\]
\(\Rightarrow 33 .\) 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.]
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 \(p=\pm \infty . ]\)
Do Problem 34 by considering \(\underline{\lim}\) and \(\overline{\lim}\) as in Problem 32.
\(\left[\text { Hint: Replace } \frac{c}{n} \text { by } \frac{c}{b_{n}}, \text { where } b_{n}=\sum_{i=1}^{n} y_{i} \rightarrow+\infty .\right]\)
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. \(]\)
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
\[
u_{n}=(m+1) S_{m n}-n^{m+1} \text { and } v_{n}=n^{m}(m+1) . ]
\]
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 \(q=\frac{1}{2}(p+q) . ]\)
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.]
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
\[
\overline{x}_{n} \cdot \overline{y}_{n} \rightarrow \overline{q} \cdot \overline{r}.
\]