# Understanding these basic definition and statement~

• Let $\left(a_{k}\right)$ be a sequence of real numbers. We use the notation $s_{n}=\sum_{k=0}^{n} a_{k}$ to denote the $n$th partial sum of the infinite series $s_{\infty}=\sum_{k=0}^{\infty} a_{k} .$ If the sequence of partial sums $\left(s_{n}\right)$ converges to a real number $s$, we say that the series $\sum_{k} a_{k}$ is convergent and we write $s=\sum_{k=0}^{\infty} a_{k} . \mathrm{A}$ series that is not convergent is called divergent.
• An infinite series $\sum_{k=0}^{\infty} a_{k}$ is said to converge absolutely if $\sum_{k=0}^{\infty}\left|a_{k}\right|$ converges. If a series converges absolutely, then it converges. Furthermore, an absolutely convergent series converges to the same sum in whatever order the terms are taken.

If $\sum^{\infty} a_{k}$ converges but $\sum_{k}^{\infty}\left|a_{k}\right|$ diverges, then we say $\sum^{\infty} a_{k}$ converges conditionally. Any conditionally convergent series can be rearranged to obtain a series which converges to any given sum or diverges to $\infty$ or $-\infty .$

• Ratio Test: Given a series $\sum_{k=1}^{\infty} a_{k}$ with $a_{k} \neq 0$, if $a_{k}$ satisfies
$$\lim {k \rightarrow \infty}\left|\frac{a{k+1}}{a_{k}}\right|=r<1$$
then the series converges absolutely.
• Root Test: Let $\sum_{k=1}^{\infty} a_{k}$ be a series and
$$\alpha:=\lim {k \rightarrow \infty} \sqrt[k]{\left|a{k}\right|}$$
Then the following hold.
a) $\sum_{k=1}^{\infty} a_{k}$ converges absolutely if $\alpha<1$. b) $\sum_{k=1}^{\infty} a_{k}$ diverges if $\alpha>1$.
For $\alpha=1$ both convergence and divergence of $\sum_{k=1}^{\infty} a_{k}$ are possible.
• Cauchy Criterion for Series: The series $\sum^{\infty} a_{k}$ converges if and only if given $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that whenever $n>m \geq N$ it follows that
$$\left|a_{m+1}+a_{m+2}+\cdots+a_{n}\right|<\varepsilon$$
• Comparison Test: Assume $\left(a_{k}\right)$ and $\left(b_{k}\right)$ are sequences satisfying $0 \leq a_{k} \leq b_{k}$ for all $k \in \mathbb{N}$.
a) If $\sum^{\infty} b_{k}$ converges, then $\sum^{\infty} a_{k}$ converges.
b) If $\sum_{k=1}^{\infty} a_{k}$ diverges, then $\sum_{k=1}^{\infty} b_{k}$ diverges.
• Geometric Series: A series is called geometric if it is of the form
$$\sum_{k=1}^{\infty} a r^{k}=a+a r+a r^{2}+a r^{3} \cdots$$
and
$$\sum_{k=1}^{\infty} a r^{k}=\frac{a}{1-r}$$
if and only if $|r|<1 .$ In case $r=1$ and $a \neq 0$, the series diverges.
• Alternating Series Test: Let $\left(a_{n}\right)$ be a sequence satisfying
a) $a_{1} \geq a_{2} \geq a_{3} \geq \cdots \geq a_{n} \geq a_{n+1} \geq \cdots$ and
b) $\left(a_{n}\right) \rightarrow 0$.
Then the alternating series $\sum_{n=1}^{\infty}(-1)^{n+1} a_{n}$ converges.
• Let $\sum_{k=1}^{\infty} a_{k}$ be a series. A rearrangement is the series $\sum_{k=1}^{\infty} a_{\sigma(k)}$ where $\sigma$ is a permutation of ${1,2,3, \ldots} .$ The summands of the rearrangement $\sum_{k=1}^{\infty} a_{\sigma(k)}$ are the same as those of the original series, but they occur in different order. If $\sigma$ is a permutation of $\mathbb{N}$ with $\sigma(k)=k$ for almost all $k \in \mathbb{N}$, then $\sum_{k=1}^{\infty} a_{k}$ and $\sum_{k=1}^{\infty} a_{\sigma(k)}$ have the same convergence behavior, and their values are equal if the series converge. For a permutation $\sigma(k) \neq k$ for infinitely many $k \in \mathbb{N}$, this may not be true.
• If $\sum_{k=1}^{\infty} a_{k}$ converges absolutely, then any rearrangement of this series converges to the same limit.

# Try the following problem to test yourself！

Problem 1.

Show that the Euler’s series $\sum_{k=1}^{\infty} \frac{1}{k^{2}}$ converges. Find the sum of the series.

Proof .

Let us use the substitution $u=\pi-x .$ Then we have
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\int_{\pi}^{0}-(\pi-u) e^{\sin (\pi-u)} d u=\int_{0}^{\pi}(\pi-u) e^{\sin (u)} d u$$
which implies
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\pi \int_{0}^{\pi} e^{\sin (u)} d u-\int_{0}^{\pi} u e^{\sin (u)} d u$$
Since
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\int_{0}^{\pi} u e^{\sin (u)} d u$$
we get

Because the terms in the sum are all positive, the sequence of partial sums given by
$$s_{n}=1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^{2}}$$
is increasing. To find an upper bound for $s_{n}$, observe
\begin{aligned} s_{n} &=1+\frac{1}{2 \cdot 2}+\frac{1}{3 \cdot 3}+\cdots+\frac{1}{n \cdot n} \ &<1+\frac{1}{2 \cdot 1}+\frac{1}{3 \cdot 2}+\cdots+\frac{1}{n \cdot(n-1)} \ &=1+\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3 \cdot 2}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right) \ &=1+1-\frac{1}{n} \ &<2 . \end{aligned}

Thus 2 is an upper bound for the sequence of partial sums, so by the Monotone Convergence Theorem, $\sum_{k=1}^{\infty} \frac{1}{k^{2}}$ converges to a limit less than 2 . Next, we claim that $\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}$. This can be shown by using the well-known clever trick of evaluating the double integral
$$I=\int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y} d x d y$$
and we evaluate $I$ in two different ways. First notice
$$\frac{1}{1-x y}=\sum_{k=0}^{\infty}(x y)^{k},$$
therefore
\begin{aligned} I &=\int_{0}^{1} \int_{0}^{1} \sum_{k=0}^{\infty}(x y)^{k} d x d y \ &=\sum_{k=0}^{\infty} \int_{0}^{1} \int_{0}^{1}(x y)^{k} d x d y \ &=\sum_{k=0}^{\infty}\left(\int_{0}^{1} x^{k} d x\right)\left(\int_{0}^{1} y^{k} d y\right) \ &=\sum_{k=0}^{\infty} \frac{1}{(k+1)^{2}} \ &=\sum_{k=1}^{\infty} \frac{1}{k^{2}} \end{aligned}

The second way to evaluate $I$ comes from a change of variables. Let
$$u=\frac{x+y}{2} \quad \text { and } \quad v=\frac{y-x}{2}$$
or equivalently
$$x=u-v \quad \text { and } \quad y=u+v$$
Given this transformation,
$$\frac{1}{1-x y}=\frac{1}{1-\left(u^{2}+v^{2}\right)}$$
and using the change of variables formula we obtain
$$I=\iint f(x, y) d x d y=\iint f(x(u, v), y(u, v))\left|\frac{d(x, y)}{d(u, v)}\right| d u d v$$
where
$$\left|\frac{d(x, y)}{d(u, v)}\right|=2$$

Since the function to be integrated and the domain in the $u v$-plane are symmetric with respect to the $u$-axis, we can split the integral into two parts as such,
\begin{aligned} I &=4 \int_{0}^{1 / 2}\left(\int_{0}^{u} \frac{d v}{1-u^{2}+v^{2}}\right) d u+4 \int_{1 / 2}^{1}\left(\int_{0}^{1-u} \frac{d v}{1-u^{2}+v^{2}}\right) d u \ &=4 \int_{0}^{1 / 2} \frac{1}{\sqrt{1-u^{2}}} \arctan \left(\frac{u}{\sqrt{1-u^{2}}}\right) d u+4 \int_{1 / 2}^{1} \frac{1}{\sqrt{1-u^{2}}} \arctan \left(\frac{1-u}{\sqrt{1-u^{2}}}\right) d u . \end{aligned}
Now, observe that if we set
$$k(u)=\arctan \left(\frac{u}{\sqrt{1-u^{2}}}\right) \text { and } h(u)=\arctan \left(\frac{1-u}{\sqrt{1-u^{2}}}\right),$$
then we obtain the derivatives
$$k^{\prime}(u)=\frac{1}{\sqrt{u^{2}}} \text { and } h^{\prime}(u)=-\frac{1}{2} \frac{1-u}{\sqrt{1-u^{2}}} \text {. }$$
This yields
\begin{aligned} I &=4 \int_{0}^{1 / 2} k^{\prime}(u) k(u) d u+4 \int_{1 / 2}^{1}-2 h^{\prime}(u) h(u) d u \ &=\left.2(k(u))^{2}\right|{0} ^{1 / 2}-\left.4(h(u))^{2}\right|{1 / 2} ^{1} \ &=2(k(1 / 2))^{2}-2(k(0))^{2}–4(h(1))^{2}+4(h(1 / 2))^{2} \ &=2\left(\frac{\pi}{6}\right)^{2}-0+0+4\left(\frac{\pi}{6}\right)^{2} \ &=\left(\frac{\pi}{6}\right)^{2} . \end{aligned}

Problem 2.

Let $T:(C[0,1],|\cdot|) \rightarrow(C[0,1],|\cdot|)$ be a function defined as
$$T f(x)=\int_{0}^{x} f(t) d t$$
where by $C[0,1]$ we mean the vector space of all continuous real-valued functions defined on $[0,1]$ and $|f|=\sup _ Discuss the convergence or divergence of$\sum x_{n}$where $$x_{n}=\frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots(2 n)}$$ Proof . Note that $$x_{n}=\frac{(2 n) !}{\left(2^{n} n !\right)^{2}}=\frac{(2 n) !}{2^{2 n}(n !)^{2}}$$ Let us first remark that$\lim {n \rightarrow \infty} x{n}=0 .$Indeed, notice that$0<x_{n}<1$for any$n \geq 1$. Since$x_{n+1}=\frac{2 n+1}{2 n+2} x_{n}$, we have$x_{n+1}<x_{n}$for any$n \geq 1$, which implies that$\left{x_{n}\right}$is decreasing. So$\lim {n \rightarrow \infty} x{n}=l$exists. Define $$y_{n}=\frac{2 \cdot 4 \cdots(2 n)}{3 \cdot 5 \cdots(2 n+1)}$$ Since$4 n^{2}-1<4 n^{2}$we get$\frac{2 n-1}{2 n}<\frac{2 n}{2 n+1}$, for any$n \geq 1 .$This obviously implies$x_{n}<y_{n}$for any$n \geq 1 .$Since$x_{n} y_{n}=\frac{1}{2 n+1}$we deduce that$x_{n}^{2}<x_{n} y_{n}=\frac{1}{2 n+1}$for any$n \geq 1 .$Hence$\lim {n \rightarrow \infty} x{n}^{2}=0$which implies$\lim {n \rightarrow \infty} x{n}=0 .$This conclusion may suggest that the series$\sum x_{n}$is convergent. But since $$\frac{x_{n+1}}{x_{n}}=\frac{2 n+1}{2 n+2}$$ we have $$n\left(1-\frac{x_{n+1}}{x_{n}}\right)=\frac{n}{2 n+2}$$ Hence$\lim {n \rightarrow \infty} n\left(1-\frac{x{n+1}}{x_{n}}\right)=\frac{1}{2}<1$, then$\sum x_{n}$is divergent based on the previous problem. Note that the ratio test does not help since$\lim {n \rightarrow \infty} \frac{x{n+1}}{x_{n}}=1 .$In fact, the root test will also be not conclusive. Problem 3. Given two series$\sum x_{n}$and$\sum y_{n}$, define$z_{n}=\sum_{k=0}^{n} x_{k} y_{n-k} .$Suppose that$\sum x_{n}$and$\sum y_{n}$are absolutely convergent. Show that$\sum z_{n}$is absolutely convergent, and $$\sum z_{n}=\sum x_{n} \cdot \sum y_{n} .$$ Proof . We show that the series $$x_{0} y_{0}+x_{0} y_{1}+x_{1} y_{1}+x_{1} y_{0}+x_{0} y_{2}+x_{1} y_{2}+x_{2} y_{2}+x_{2} y_{1}+x_{2} y_{0}+\cdots$$ converges absolutely. In fact, the sum of the first$(n+1)^{2}$terms of the series$(8.4)\left|x_{0} y_{0}\right|+\left|x_{0} y_{1}\right|+\left|x_{1} y_{1}\right|+\left|x_{1} y_{0}\right|+\left|x_{0} y_{2}\right|+\left|x_{1} y_{2}\right|+\left|x_{2} y_{2}\right|+\left|x_{2} y_{1}\right|+\left|x_{2} y_{0}\right|+\cdots$is$\sum_{k=0}^{n}\left|x_{k}\right| \cdot \sum_{k=0}^{n}\left|y_{k}\right|$, which converges to$\sum_{k=0}^{\infty}\left|x_{k}\right| \cdot \sum_{k=0}^{\infty}\left|y_{k}\right|$. Hence the sequence of partial sums of$(8.4)$has a convergent subsequence. But all the terms of the series are positive, so the sequence of all partial sums is increasing, and bounded above by$\sum_{k=0}^{\infty}\left|x_{k}\right| \cdot \sum_{k=0}^{\infty}\left|y_{k}\right|$, so it also converges, to the same limit. Thus (8.3) converges absolutely. The same argument as above, considering sums of the first$(n+1)^{2}$terms of the series$(8.3)$, shows that the sum of this series is$\sum x_{n} \cdot \sum y_{n}$But $$\sum z_{n}=x_{0} y_{0}+\left(x_{0} y_{1}+x_{1} y_{0}\right)+\left(x_{0} y_{2}+x_{1} y_{1}+x_{2} y_{0}\right)+\cdots$$ is a rearrangement of$(8.3)$, so by the Rearrangement Theorem it converges to the same limit. Note: You might like to think about the above proof by considering an “infinite matrix” in which the$(i, j)$term is$x_{i} y_{i}$:$\begin{array}{lllll}x_{0} y_{0} & x_{0} y_{1} & x_{0} y_{2} & x_{0} y_{3} & \cdots \ x_{1} y_{0} & x_{1} y_{1} & x_{1} y_{2} & x_{1} y_{3} & \cdots \ x_{2} u_{0} & x_{2} u_{1} & x_{2} u_{2} & x_{2} y_{2} & \cdots\end{array}$$$\cdots$$$0 \quad x_{3}\$
$$\begin{array}{r} \cdots \ \ddots \end{array}$$

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