城南讲马堂

陶哲轩实分析7.5及习题-Analysis I 7.5

2019年4月26日2020年2月15日christangdt8条评论

两个很重要的test：root test和ratio test，判别级数收敛性的两个重要方法，其中root test比ratio test更聚集一些（Lemma 7.5.2）

Exercise 7.5.1. Prove the first inequality in Lemma 7.5.2.
Solution: To prove

$\liminf\limits_{n\to \infty}\frac{c_{n+1}}{c_n} \leq \liminf\limits_{n\to \infty}c_n^{1/n}$

We notice $c_{n+1}/c_n\geq 0$ for all $n$ , so

$L:=\liminf\limits_{n\to \infty} \frac{c_{n+1}}{c_n} \geq 0$

If $L=0$ , then the inequality is true since $c_n>0 \implies c_n^{1/n}>0 \implies \liminf_{n\to \infty}c_n^{1/n} \geq 0$ . Now we suppose $L>0$ .
For $\forall \varepsilon >0$ , such that $L-\varepsilon >0$ , we can find a $N\geq m$ such that

$\dfrac{c_{n+1}}{c_n}>L-\varepsilon ,\quad \forall n\geq N$

From induction we know that

$c_n>(L-\varepsilon )^{n-N} c_N,\quad \forall n\geq N$

We let $A=(L-\varepsilon )^{-N} c_N$ , then $c_n>A(L-\varepsilon )^n \implies c_n^{1/n}>A^{1/n} (L-\varepsilon )$ , use limit laws we have

$\liminf\limits_{n\to \infty}c_n^{\frac{1}{n}} \geq \liminf\limits_{n\to \infty}A^{\frac{1}{n}} (L-\varepsilon ) =\lim\limits_{n\to \infty}A^{\frac{1}{n}} (L-\varepsilon )=L-\varepsilon$

Since $\varepsilon$ can be arbitrary small (as long as $\varepsilon <L$ ), we must have $\liminf_{n\to \infty}c_n^{1/n} \geq L$ .

$\blacksquare$

Exercise 7.5.2. Let $x$ be a real number with $|x|<1$ , and $q$ be a real number. Show that the series $\sum_{n=1}^{\infty}n^qx^n$ is absolutely convergent, and that $\lim_{n\to\infty}n^qx^n=0$ .
Solution: If $q=0$ , then $\sum_{n=1}^{\infty}n^q x^n =\sum_{n=1}^{\infty}x^n$ and is absolutely convergent. Now first suppose $q>0$ .
We use Ratio test, since $a_n=n^q x^n$ , so that

$\dfrac{|a_{n+1} |}{|a_n |} =\dfrac{|(n+1)^q x^{n+1} |}{|n^q x^n |} =\left(\dfrac{n+1}{n}\right)^q |x|$

Now as $|x|<1$ , we know that

$|x|<\dfrac{|x|+1}{2}<1 \implies \dfrac{2}{|x|+1}>1 \implies \left(\dfrac{2}{|x|+1}\right)^{1/q}>1$

Since

$\lim\limits_{n\to \infty}\frac{n+1}{n}=1$

We know that there’s $N\geq 1$ , such that $\forall n\geq N$ :

$1<\dfrac{n+1}{n}\leq \left(\dfrac{2}{|x|+1}\right)^{1/q} \implies \left(\dfrac{n+1}{n}\right)^q |x|\leq \dfrac{2|x|}{|x|+1}$

Thus

$\limsup\limits_{n\to \infty} \dfrac{|a_{n+1}|}{|a_n |} =\limsup\limits_{n\to \infty}\left(\dfrac{n+1}{n}\right)^q |x| \leq \dfrac{2|x|}{|x|+1}<1$

This shows that $\sum_{n=1}^{\infty}n^q x^n$ is absolutely convergent.
Next suppose $q<0$ , then $-q>0$ , so

$\dfrac{|a_{n+1}|}{|a_n |} =\left(\dfrac{n+1}{n}\right)^q |x|=\left(\dfrac{n}{n+1}\right)^{-q} |x|<1^{-q} |x|=|x|$

Thus

$\limsup\limits_{n\to \infty}\dfrac{|a_{n+1}|}{|a_n |} \leq |x|<1$

This shows that $\sum_{n=1}^{\infty}n^q x^n$ is absolutely convergent.
By the zero test, $\lim_{n\to \infty}n^q x^n =0$ .

$\blacksquare$

Exercise 7.5.3. Give an example of a divergent series $\sum_{n=1}^{\infty}a_n$ of positive numbers $a_n$ such that $\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}a_n^{1/n}=1$ , and give an example of a convergent series $\sum_{n=1}^{\infty}b_n$ of positive numbers $b_n$ such that $\lim_{n\to\infty}b_{n+1}/b_n=\lim_{n\to\infty}b_n^{1/n}=1$ . This shows that the ratio and root tests can be inconclusive even when the summands are positive and all the limits converge.
Solution: We let $a_n=1/n$ and $b_n=1/n^2$ , then $\sum_{n=1}^{\infty}a_n$ is divergent and $\sum_{n=1}^{\infty}b_n$ is convergent by Corollary 7.3.7. Next we use Proposition 7.5.4 and limit laws we can get

$\dfrac{a_{n+1}}{a_n} =\dfrac{n}{n+1} \implies \lim\limits_{n\to \infty}\dfrac{a_{n+1}}{a_n}=1 \\a_n^{1/n}=\left( \dfrac{1}{n}\right)^{1/n} \implies \lim\limits_{n\to \infty}a_n^{1/n}=1 \\ \dfrac{b_{n+1}}{b_n} =\dfrac{n^2}{(n+1)^2} \implies \lim\limits_{n\to \infty}\dfrac{b_{n+1}}{b_n}=1\times 1=1 \\b_n^{1/n}=\left( \dfrac{1}{n^2}\right)^{1/n} \implies \lim\limits_{n\to \infty}b_n^{1/n}=1^2=1$

$\blacksquare$

陶哲轩实分析7.4及习题-Analysis I 7.4

2019年4月26日2019年12月13日christangdt5条评论

级数重排对绝对收敛是可以的，但条件级数重排可以收敛到任意其他值或者发散……

Exercise 7.4.1. Let $\sum_{n=0}^{\infty}a_n$ be an absolutely convergent series of real numbers. Let $f:\mathbf N \to\mathbf N$ be an increasing function(i.e., $f(n+1)>f(n)$ for all $n\in \mathbf N$ ). Show that $\sum_{n=0}^{\infty}a_{f(n)}$ is also an absolutely convergent series.
Solution: We consider $\sum_{n=0}^{\infty}|a_{f(n)}|$ , and prove it is convergent. Suppose $\sum_{n=0}^{\infty}|a_n| =L$ .
Let $T_N=\sum_{n=0}^N|a_{f(n)}|$ , since the sequence $\big(f(n)\big)_{n=0}^N$ is finite, it must be bounded by $M$ , thus the set $\{n\in \mathbf N:0\leq n\leq f(N)\}\subseteq \{n\in \mathbf N:0\leq n\leq M\}$ , so we have

$T_N=\sum\limits_{n=0}^N|a_{f(n)} | \leq \sum\limits_{n=0}^M|a_n | \leq \sum\limits_{n=0}^{\infty}|a_n | =L$

So $(T_N )_{N=0}^{\infty}$ is bounded above by $L$ , thus it must be convergent.

$\blacksquare$

陶哲轩实分析7.3及习题-Analysis I 7.3

2019年4月26日2019年12月13日christangdt留下评论

处理非负实数的和，与绝对收敛的处理类似，和Cauchy序列类比，得到关于 $1/n^q$ 序列收敛与否的重要结论。

Exercise 7.3.1. Use Proposition 7.3.1 to prove Corollary 7.3.2.
Solution: That $\sum_{n=m}^{\infty}b_n$ is convergent means $\sum_{n=m}^{\infty}b_n =L\in \mathbf R$ , and $|a_n|\leq b_n,\forall n\geq m$ means the partial sums $\sum_{n=m}^N|a_n| \leq \sum_{n=m}^Nb_n ,\forall N\geq m$ , then use the limit laws and Proposition 7.2.9 we get the conclusion.

$\blacksquare$

Exercise 7.3.2. Prove Lemma 7.3.3.
Solution: We let $S_N=\sum_{n=0}^Nx^n$ , we use induction to show that $S_N=(1-x^{N+1})/(1-x)$ :
When $N=0$ , we have $\sum_{n=0}^0x^n =x^0=1$ and $(1-x^{0+1})/(1-x)=1$ .
Now suppose $S_N=(1-x^{N+1})/(1-x)$ , we have

$S_{N+1}=\sum\limits_{n=0}^{N+1}x^n =\sum\limits_{n=0}^Nx^n +x^{N+1}=\dfrac{1-x^{N+1}}{1-x}+x^{N+1}=\dfrac{1-x^{N+2}}{1-x}$

Next, we know from Lemma 6.5.2 that

$\lim\limits_{n\to {\infty}}x^n=\begin{cases}0,&|x|<1\\1,&x=1\\ \text{diverges},&x=-1\text{ or }|x|>1\end{cases}$

So if $|x|<1$ , then when $N\to {\infty}$ , we have $S_N\to 1/(1-x)$ , in particular, we can show that

$S_N'=\sum\limits_{n=0}^N|x|^n =\dfrac{1-|x|^{N+1}}{1-|x|}$

And $\lim_{n\to {\infty}}|x|^n=0$ since $\big||x|\big|=|x|<1$ , so the series is absolutely convergent.
If $|x|\geq 1$ , we can know that $x^n\nrightarrow 0$ , thus use Zero test we can know the series is divergent.

$\blacksquare$

Exercise 7.3.3. Let $\sum_{n=0}^{\infty}a_n$ be an absolutely convergent series of real numbers such that $\sum_{n=0}^{\infty}|a_n|=0$ . Show that $a_n=0$ for every natural number $n$ .
Solution: We suppose $a_m=k\neq 0$ for some $m\in \mathbf N$ , then $|a_m|=|k|>0$ , thus

$\sum\limits_{n=0}^m|a_n|\geq |a_m|>0$

Which means

$0=\left(\sum\limits_{n=0}^{\infty}|a_n| \right)\geq \left(\sum\limits_{n=0}^m|a_n| \right)\geq |a_m|>0$

This is a contradiction.

$\blacksquare$

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