the sequence of prime numbers, and we have reason to believe that it is a
mystery into which the human mind will never penetrate.
Understanding these basic definition and statement~
- A sequence is a function whose domain is the set $\mathbb{N}$ of natural numbers.
- A sequence $\left{x_{n}\right}$ is said to converge to a real number $x$, provided that for each $\varepsilon>0$ there exists an integer $N$ such that $n \geq N$ implies that $\left|x_{n}-x\right|<\varepsilon$ In this case we also say that $\left{x_{n}\right}$ converges to $x$, or $x$ is the limit of $\left{x_{n}\right}$, and we write $x_{n} \rightarrow x$, or $\lim {n \Rightarrow \infty} x{n}=x .$ If $\left{x_{n}\right}$ does not converge, it is said to diverge.
- A sequence $\left{x_{n}\right}$ is said to be bounded if the range $\left{x_{n}: n \in \mathbb{N}\right}$ is a bounded set, that is, if there exists $M \geq 0$ such that $\left|x_{n}\right| \leq M$ for all $n \in \mathbb{N}$.
- Bolzano-Weierstrass Theorem: Every bounded sequence has a convergent subsequence.
- Let $\left{x_{n}\right}_{n=1}^{\infty}$ be a sequence and for each $n \in \mathbb{N}$, set
$$
y_{n}=\sup \left{x_{k}: k \geq n\right} .
$$
The limit superior of $\left{x_{n}\right}$, denoted by $\lim \sup \left{x_{n}\right}$ or $\varlimsup_{n}\left{x_{n}\right}$, is defined by
$$
\varlimsup_{n}\left{x_{n}\right}=\inf \left{y_{n}: n \in \mathbb{N}\right}=\inf \left{x: x=\sup \left{x_{k}: k \geq n\right} \text { for some } n \in \mathbb{N}\right}
$$
provided that the quantity on the right exists. Likewise we define the limit inferior by
$$
\varliminf_{{}\left{x_{n}\right}=\sup \left{x: x=\inf \left{x_{k}: k \geq n\right} \text { for some } n \in \mathbb{N}\right}
$$
It is well known that if $\left{x_{n}\right}$ is a sequence, then $\left{x_{n}\right}$ has a limit if and only if the limit superior and the limit inferior exist and are equal.
- A sequence $\left{x_{n}\right}$ of real numbers is said to be a Cauchy sequence if for every $\varepsilon>0$, there is an integer $N$ such that
$$
\left|x_{n}-x_{m}\right|<\varepsilon \text { if } n \geq N \text { and } m \geq N
$$
Let $\left{x_{n}\right}_{n=1}^{\infty}$ be a sequence and let $\left{n_{k}\right}_{k=1}^{\infty}$ be any sequence of natural numbers such that $n_{1}<n_{2}<n_{3}<\ldots$ The sequence $\left{x_{n_{k}}\right}_{k=1}^{\infty}$ is called a subsequence of $\left{x_{n}\right}_{n=1}^{\infty}$.
Try the following problem to test yourself！
Let $\left{x_{n}\right}$ and $\left{y_{n}\right}$ be two real sequences such that
(a) $x_{n} \leq y_{n}$ for all $n$;
(b) $\left{x_{n}\right}$ is increasing;
(c) $\left{y_{n}\right}$ is decreasing.
Show that $\left{x_{n}\right}$ and $\left{y_{n}\right}$ are convergent and
$$
\lim {n \rightarrow \infty} x{n} \leq \lim {n \rightarrow \infty} y{n} \text {. }
$$
When do we have equality of the limits?
Since $\left{y_{n}\right}$ is decreasing, we have $y_{n} \leq y_{1}$ for $n \geq 1 .$ So for any $n \geq 1$ we have $x_{n} \leq y_{n} \leq y_{1}$. This implies that $\left{x_{n}\right}$ is bounded above. Since it is increasing it converges. Similar argument shows that $\left{y_{n}\right}$ is bounded below and therefore converges as well. From (a) we get the desired inequality on the limits. In order to have the equality of the limits we must have $\lim {n \rightarrow \infty} y{n}-x_{n}=0$. This result is useful when dealing with nested intervals in $\mathbb{R}$ and alternating real series.
Discuss the convergence or divergence of
$$
x_{n}=\frac{[\alpha]+[2 \alpha]+\cdots+[n \alpha]}{n^{2}},
$$
where $[x]$ denotes the greatest integer less than or equal to the real number $x$, and $\alpha$ is an arbitrary real number.
By definition of the greatest integer function $[\cdot]$, we have
$$
[x] \leq x<[x]+1
$$
for any real number $x$. This will easily imply $x-1<[x] \leq x$. So
$$
\frac{(\alpha-1)+(2 \alpha-1)+\cdots+(n \alpha-1)}{n^{2}}<\frac{[\alpha]+[2 \alpha]+\cdots+[n \alpha]}{n^{2}} \leq \frac{\alpha+2 \alpha+\cdots+n \alpha}{n^{2}}
$$
or
$$
\frac{(1+2+\cdots+n) \alpha-n}{n^{2}}<\frac{[\alpha]+[2 \alpha]+\cdots+[n \alpha]}{n^{2}} \leq \frac{(1+2+\cdots+n) \alpha}{n^{2}} .
$$
The algebraic identity $1+2+\cdots+m=\frac{m(m+1)}{2}$ for any natural number $m \geq 1$ gives
$$
\frac{\frac{n(n+1)}{2} \alpha-n}{n^{2}}<\frac{[\alpha]+[2 \alpha]+\cdots+[n \alpha]}{n^{2}} \leq \frac{\frac{n(n+1)}{2} \alpha}{n^{2}}
$$
or
$$
\frac{(n+1) \alpha}{2 n}-\frac{1}{n}<\frac{[\alpha]+[2 \alpha]+\cdots+[n \alpha]}{n^{2}} \leq \frac{(n+1) \alpha}{2 n} .
$$
Since
$$
\lim {n \rightarrow \infty} \frac{(n+1) \alpha}{2 n}-\frac{1}{n}=\frac{\alpha}{2} \text { and } \lim {n \rightarrow \infty} \frac{(n+1) \alpha}{2 n}=\frac{\alpha}{2},
$$
the Squeeze Theorem implies $\lim {n \rightarrow \infty} x{n}=\frac{\alpha}{2}$.
Show that $\left{x_{n}\right}$ defined by
$$
x_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}
$$
is divergent.
We have
$$
x_{2 n}-x_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}
$$
for any $n \geq 1$. So
$$
\frac{1}{n+n}+\frac{1}{n+n}+\cdots+\frac{1}{2 n} \leq x_{2 n}-x_{n}
$$
or $\frac{1}{2} \leq x_{2 n}-x_{n} .$ This clearly implies that $\left{x_{n}\right}$ fails to be Cauchy. Therefore it diverges.
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Fourier analysis代写
离散数学代写
Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析
