- Show directly from the definition of “convergent sequence” that if the sequences $a_n$ and $b_n$ both converge, and if $a_n\leq b_n$ for all $n\in\N$, then $\lim_{n\to\infty}a_n \leq \lim_{n\to\infty}b_n$
- Prove that if the sequences $a_n$ and $b_n$ are bounded, and if $a_n\leq b_n$ for all $n\in\N$ then \[ \liminf_{n\to\infty} a_n \leq \liminf_{n\to\infty} b_n \quad\text{ and }\quad \limsup_{n\to\infty} a_n\leq \limsup_{n\to\infty} b_n. \]
- Use part b of this problem to prove the fat sandwich theorem.
- Let $x_n\in\R$ be a bounded sequence and assume $x_{n_k}$ is a convergent subsequence. Prove that \[ \liminf _{n\to\infty} x_n \leq \lim_{k\to\infty} x_{n_k} \leq \limsup_{n\to\infty} x_n. \]
- Show that if the series $\sum_{n=1}^\infty a_n$ converges, and if $a_n\geq0$ for all $n\in\N$, then $\sum_{n=1}^\infty a_n^2$ also converges.
- Find a sequence $a_n$ for which $\sum_{n=1}^\infty a_n$ does not converge, but for which $\sum_{n=1}^\infty a_n^2$ converges.
- Find a sequence $a_n$ for which $\sum_{n=1}^\infty a_n^2$ diverges even though $\sum_{n=1}^\infty a_n$ does converge.
a. Hint: If the series $\sum a_n$ converges, then $a_n\to 0$;
if $a_n\to 0$ then the sequence $a_n$ is bounded, i.e. there is some
$M$ with $a_n\leq M$ for all $n\in\N$. If $0\leq a_n \leq M$ then $a_n^2 =a_n
\times a_n \leq Ma_n$.
b. Hint: consider $\sum_{n=1}^\infty n^{-p}$ for suitably
chosen $p$.
c. Hint: consider $\sum_{n=1}^\infty (-1)^{n+1} n^{-p}$ for
suitably chosen $p$.
Hint: either remember the Cauchy–Schwarz inequality, or use that
$|2ab|\leq a^2 + b^2$ for all $a,b\in\R$ (which follows from expanding
$(a-b)^2\geq 0$).
- Show that $a_n\to 0$, directly from the definition.
Given $\varepsilon\gt0$ choose an integer $N$ such that $N\gt 1/\varepsilon$. Then for all $n\ge N$ one has $0\lt a_n \le 1/N \lt \varepsilon$.
- Show that if a sequence $x_n$ converges, then any
subsequence of $x_n$ also converges and has the same limit.
Let $y_k = x_{n_k}$ be a subsequence of $x_n$, and let $L$ be the limit of $x_n$. In particular this implies that $n_k\ge k$ for all $k$. Given $\varepsilon\gt0$ we choose $N\in\N$ such that $|x_n-L|\lt \varepsilon$ for all $n\ge N$. Then, if $k\ge N$ we have $n_k\ge k\ge N$ and thus $|y_k-L| = |x_{n_k}-L| \lt \varepsilon$.
- Show that $b_n=n^{-2}\to 0$ without using the definition of
convergence.
$b_n = a_{n^2}$, so $b_n$ is a subsequence of the sequence $a_n=1/n$. Therefore it converges, and it has the same limit: $\lim_{n\to\infty} 1/n^2 = 0$.
Let $x_n\to p$, i.e. let $p$ be the
limit of the sequence $x_n$. Choose $\varepsilon=1$. There
is an $N\in\N$ such that for all $n\ge N$ on has $d(x_n, p)\lt
\varepsilon=1$. Thus, if
\[
R=\max\left\{1, d(x_1, p), \dots, d(x_{N-1}, p)\right\},
\]
then all $x_k$ satisfy $d(x_k, p)\le R$.
In order to reach a contradiction, assume that $x\not\in C$.
The complement $C^c$ of $C$ is open in $X$, so there is an
$\varepsilon\gt0$ such that $B_\varepsilon(x)\subset C^c$.
Since $x_n\to x$, there is an $N_\varepsilon\in\N$ for such
that $x_n\in B_\varepsilon(x)$ for all $n\ge N$. But then we
would have $x_n\not\in C$ for $n\ge N$, which is a
contradiction. We conclude that $x\in C$ after all.
The sequences $x_n\in\R^k$ and $y_n\in\R^k$ converge, so they
are bounded, i.e. there is an $R\gt0$ such that $\|x_n\|\le
R$, and $\|y_n\|\le R$ for all $n\in\N$. We also have
\begin{align*}
\left|x_n\cdot y_n - x\cdot y\right|
&=\left| x_n\cdot y_n - x_n\cdot y + x_n\cdot y - x\cdot y\right| \\
&\le \left| x_n\cdot y_n - x_n\cdot y\right| + \left|x_n\cdot y - x\cdot y\right|
& \text{triangle }\le\\
&\le \left| x_n\cdot (y_n - y)\right| + \left|(x_n - x)\cdot y\right| \\
&\le \| x_n\|\,\| y_n - y\| + \|x_n - x\|\,\| y\|
& \text{Cauchy }\le \\
&\le R\| y_n - y\| + R \|x_n - x\|
\end{align*}
Since $x_n$ and $y_n$ converge we have
\[
\lim_{n\to\infty} R\| y_n - y\| + R \|x_n - x\| = 0
\]
and hence
\[
\lim_{n\to\infty} \left|x_n\cdot y_n - x\cdot y\right| =0.
\]
Suppose $x$ is not an upper bound for $A$.
Then there is a number $a\in A$ with $a\gt x$. Let $\varepsilon=
a-x$. Since $x_n\to X$, there is an $N$ such that for all $n\ge
N$ one has $|x_n-x|\lt \varepsilon$. This implies, for $n\ge N$,
that $x_n \lt x+\varepsilon = a$. Hence $x_n$ is not an upper
bound for $A$ if $n\ge N$, contradicting what was given.
Let $\varepsilon\gt0$ and $n\in\N$ be given.
By definition of “lim sup” we have
\[
\lim_{m\to\infty} \Bigl(\sup\bigl\{a_k : k\ge m\bigr\}\Bigr) =A.
\]
By definition of “limit” this implies that there is an
$N_\varepsilon\in\N$ such that for all $m\ge N_\varepsilon$ we have
\[
A - \varepsilon \lt
\sup\bigl\{a_k : k\ge m\bigr\}\lt
A + \varepsilon
\]
We choose $m \gt \max \{N_\varepsilon, n\}$. Then
\[
\sup\{a_k : k\ge m\} \gt A - \varepsilon.
\]
Since $\sup \{\dots\}$ is the least upper bound it follows that $A-
\varepsilon$ is not an upper bound for $\{a_k : k\ge m\}$. Thus there is an
$k\ge m$ such that $a_k \gt A-\varepsilon$. We had chosen $m\gt n$, and
thus $k\gt n$, so that we have found a $k\gt n$ for which $a_k\gt A-
\varepsilon$.
- $a_n = (-1)^n$
- $a_n = (-1)^n+ \frac 2n$
- $a_n = (-1)^n \frac {n+2}n$
- $a_n = n$
- $a_n = (-1)^n n$
- $a_n = \bigl(1 + (-1)^n\bigr) n$
In each case compute $m_n = \inf \{a_k : k\ge n\}$, and $M_n = \sup \{a_k :
k\ge n\}$. Then $\lim_{n\to\infty} m_n$ and $\lim_{n\to\infty}M_n$ are the
lim inf and lim sup, respectively.
- $a_n = (-1)^n$: $m_n=-1$ and $M_n=+1$ for all $n$, so $\liminf_{n\to\infty} a_n = -1$, while $\limsup_{n\to\infty}a_n = +1$.
- $a_n = (-1)^n+ \frac 2n$: $m_n = -1$ for all $n$. $M_{n} = 1+\frac2n$ if $n$ is even, and $M_n=M_{n+1}$ if $n$ is odd. Hence $\liminf_{n\to\infty} a_n = -1$, while $\limsup_{n\to\infty}a_n = +1$.
- $a_n = (-1)^n \frac {n+2}n$: $m_n=-\frac{n+2}n$ if $n$ is odd, and $m_{n}=m_{n+1}$ if $n$ is even, so $\liminf_{n\to\infty} a_n = -1$; $M_n=\frac{n+2}n$ if $n$ is even, and $M_{n}=M_{n+1}$ if $n$ is odd, so $\limsup_{n\to\infty} a_n = 1$.
- $a_n = n$: $m_n=n$, and $M_n=+\infty$ for all $n$. Both lim sup and lim inf are $+\infty$.
- $a_n = (-1)^n n$: $m_n=-\infty$, $M_n=+\infty$. $\liminf_{n\to\infty} a_n = -\infty$, while $\limsup_{n\to\infty}a_n = +\infty$.
- $a_n = \bigl(1 + (-1)^n\bigr) n$ : $m_n=0$, $M_n= 2n$ if $n$ is even, $M_n=M_{n+1}$ if $n$ is odd. $\liminf_{n\to\infty} a_n = 0$, while $\limsup_{n\to\infty}a_n =\infty$.
- $\sup_{n\in\N} x_n$
- all possible limits of subsequences of $x_n$
- $\limsup_{n\to\infty} x_n$
Let $x_{n}=\cos(\theta +n\frac{\pi }{3})$ where $\theta \in (0,\frac{\pi }{3})$ is some constant. For this sequence, compute:
There are only six different values in this sequence, and the sequence
repeats : $x_{n+6} = x_n$ for all $n$. You can read off the values by
drawing a unit circle.
a. $\sup_{n\in\mathbb{N}}x_{n}$
Since the sequence repeats, the supremum is the largest of the six cosines that appear in this sequence. If you work that out in more detail, then you find that $\cos \theta = \cos(\theta+2\pi)$ is the largest when $0\lt \theta\le \pi/6$, while $\cos(\theta+5\pi/3)$ is the largest if $\pi/6\le \theta\lt \pi/3$.
\[
\sup_{n\in\N}x_{n}=\left\{\begin{matrix}
\cos(\theta+\frac{5\pi}{3}),\frac{\pi}{3} \gt \theta \gt \frac{\pi}{6}\\
\cos(\theta+2\pi),0 \lt \theta\leq \frac{\pi}{6}
\end{matrix}\right.
\]
b. all possible limits of subsequences of $x_{n}$ are
\[
\cos(\theta+\frac{\pi}{3}),\quad \cos(\theta+\frac{2\pi}{3}),\quad
\cos(\theta+{\pi}),\quad \cos(\theta+\frac{4\pi}{3}),\quad
\cos(\theta+\frac{5\pi}{3}), \cos(\theta+2\pi).
\]
c. $\limsup_{n\to\infty }(x_{n})$
\[
\limsup_{n\to\infty }(x_{n})=\lim_{n\to\infty}\left \{ \sup_{k\geq n}(x_{k}) \right \}
\]
For any $n$ the least upper bound of $\{x_n, x_{n+1}, x_{n+2}, \dots\}$, i.e. $\sup_{k\ge n} x_k$, is again the largest of the six cosines you listed. So the answer is the same as in 1a:
\[
\limsup_{n\to\infty}x_{n}=\left\{\begin{matrix}
\cos(\theta+\frac{5\pi}{3}),\frac{\pi}{3}> \theta>\frac{\pi}{6}\\
\cos(\theta+2\pi),0<\theta\leq \frac{\pi}{6}
\end{matrix}\right.
\]
Assume $x_n\to p$. Let $\varepsilon\gt0$ be given.
Then there is an $N_\varepsilon\in\N$ such that for all $n\ge N_\varepsilon$
one has $d(x_n, p)\lt \varepsilon$. If $n, m\ge N_\varepsilon$, then
\[
d(x_n, x_m) \le d(x_n, p)+d(p, x_m) \le \frac \varepsilon2 +
\frac\varepsilon2 =\varepsilon.
\]
Thus $\{x_n\}$ is a Cauchy sequence.
There are many examples. Perhaps the simplest example is to
take any metric space $(X,d)$ and a convergent sequence $x_n\in X$,
$p=\lim_{n\to\infty}x_n$, and then consider the space $(Y, d)$ where
$Y=X\setminus \{p\}$, and $d$ is the same metric as on $X$. For instance,
let $Y=\R\setminus\{0\}$, with distance $d(x, y) = |x-y|$; then the sequence
$x_n = \frac1n$ is a Cauchy sequence, but it does not converge in $Y$.
The real line. Compact metric spaces are bounded, and
while $\R$ is complete, it is not bounded.
By definition
\[
\limsup_{n\to\infty}(x_n+y_n)
= \lim_{n\to\infty} \Bigl(\sup\{x_k+y_k : k\ge n\}\Bigr).
\]
For every $k\ge m$ we have
\[
x_k \le \sup\{x_l : l\ge m\}, \qquad
y_k \le \sup\{y_l : l\ge m\}.
\]
Hence, for all $k\ge m$,
\[
x_k+y_k \le \sup\{x_l : l\ge m\} + \sup\{y_l : l\ge m\}.
\]
The quantity on the right does not depend on $k$ and thus it is an upper
bound for $\{x_k+y_k : k\ge m\}$. It is therefore not smaller than the
least upper bound:
\[
\sup\{x_k+y_k : k\ge m\} \le \sup\{x_l : l\ge m\} + \sup\{y_l : l\ge m\}.
\]
If we now take the limit for $m\to\infty $ on both sides we get
\[
\limsup_{k\to\infty} x_k+y_k \le \limsup_{k\to\infty} x_k +
\limsup_{k\to\infty} y_k.
\]
There are many possibilities. Here is one:
\begin{align*}
\{x_n\} &= \{+1, 0, +1, 0, +1, 0, +1, 0, \dots \}\\
\{y_n\} &= \{-1, 0, -1, 0, -1, 0, -1, 0, \dots \}
\end{align*}
For these sequences $x_n+y_n = 0$ for all $n$, so
$\limsup_{n\to\infty}x_n+y_n =0$. But
\[
\limsup x_n = 1, \qquad \limsup y_n =0,
\]
so $\limsup(x_n+y_n) \lt \limsup x_n+ \limsup y_n$.
Abbreviate
\[
\lim_{n\to\infty} x_n = X, \qquad
\limsup_{n\to\infty}y_n = Y.
\]
By definition
\[
\limsup_{n\to\infty} \bigl(x_n +y_n\bigr)
= \lim_{n\to\infty} \Bigl\{\sup_{k\ge n} \bigl(x_n+y_n\bigr) \Bigr\}.
\]
Let $\varepsilon\gt 0$ be given. Then there is an $N_\varepsilon$ such that
for all $n\ge N_\varepsilon$ we have
\[
X-\varepsilon \lt x_n \lt X+\varepsilon,
\]
and
\[
Y - \varepsilon \lt \sup_{k\ge n} y_k \lt Y + \varepsilon.
\]
For any $k\ge N_\varepsilon$ we then have
\[
X-\varepsilon + y_k \lt x_k + y_k \lt X+\varepsilon +y_k,
\]
and thus for any $n\ge N_\varepsilon$,
\[
X-\varepsilon + \sup_{k\ge n}y_k
\lt \sup_{k\ge n} \bigl(x_k + y_k\bigr)
\lt X+\varepsilon +\sup_{k\ge n}y_k,
\]
and therefore, for all $n\ge N_\varepsilon$,
\[
X+ Y - 2\varepsilon \lt \sup_{k\ge n} \bigl(x_k + y_k\bigr) \lt X + Y + 2\varepsilon.
\]
This implies, by definition of “$\lim_{n\to \infty} (\dots)$”
that
\[
\lim_{n\to\infty} \Bigl\{\sup_{k\ge n} \bigl(x_k + y_k\bigr) \Bigr\}
= X+Y.
\]