此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。
PDF格式教材下载 Sequences and Series
本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution
Summary
- Suppose that $left(a_n ight)$ is a sequence. To say that $lim_{n o infty}a_n=L$ is to say that for every $varepsilon>0$, there is an $N > 0$, so that whenever $n>N$, we have $|a_n-L| < varepsilon$. If $lim_{n oinfty}a_n=L$ we say that the sequence converges. If there is no finite value $L$ so that $lim_{n oinfty}a_n = L$, then we say that the limit does not exist, or equivalently that the sequence diverges.
- Suppose $(a_n)$ is a sequence with initial index $N$, and suppose we have a sequence of integers $(n_i)$ so that $$N leq n_1 < n_2 < n_3 < n_4 < n_5 < cdots$$ Then the sequence $(b_i)$ given by $b_i = a_{n_i}$ is said to be a subsequence of the sequence $a_n$.
- If $(b_i)$ is a subsequence of the convergent sequence $(a_n)$, then $$lim_{i o infty} b_i = lim_{n o infty} a_n$$
- Suppose $(b_i)$ and $(c_i)$ are convergent subsequences of the sequence $(a_n)$, but $$lim_{i o infty} b_i eq lim_{i o infty} c_i.$$ Then the sequence $(a_n)$ does not converge.
- Squeeze Theorem: Suppose there is some $N$ so that for all $n > N$, it is the case that $a_n le b_n le c_n$. If $$lim_{n oinfty}a_n=lim_{n oinfty}c_n=L$$ then $lim_{n oinfty}b_n=L$.
- $$lim_{n oinfty}|a_n|=0$$ if and only if $$lim_{n oinfty}a_n=0$$
- The sequence $a_n = r^n$ converges when $-1 < r le 1$, and diverges otherwise. In symbols, $$lim_{n oinfty} r^n=egin{cases}0& mbox{if $-1 < r < 1$,} \ 1& mbox{if $r=1$, and} \ mbox{does not exist} & mbox{if $r leq -1$ or $r > 1$.} end{cases}$$
- A sequence is called increasing (or sometimes strictly increasing) if $a_n < a_{n+1}$ for all $n$. It is called non-decreasing if $a_nle a_{n+1}$ for all $n$. Similarly a sequence is decreasing (or, by some people, strictly decreasing) if $a_n > a_{n+1}$ for all $n$ and non-increasing if $a_nge a_{n+1}$ for all $n$.
- If a sequence is increasing, non-decreasing, decreasing, or non-increasing, it is said to be monotonic.
- A sequence $(a_n)$ is bounded above if there is some number $M$ so that for all $n$, we have $a_nle M$. Likewise, a sequence $(a_n)$ is bounded below if there is some number $M$ so that for every $n$, we have $a_nge M$. If a sequence is both bounded above and bounded below, the sequence is said to be bounded.
- If the sequence $a_n$ is bounded and monotonic, then $lim_{n o infty} a_n$ exists. In short, bounded monotonic sequences converge.
Exercises
1. Compute $$lim_{x oinfty} x^{1/x}$$
Solution:
$$lim_{x oinfty} x^{1/x}=lim_{x oinfty}(e^{ln x})^{1/x} =lim_{x oinfty}e^{frac{ln x}{x}}$$ By L'Hopital's rule, we have $$lim_{x oinfty}frac{ln x}{x}=lim_{x oinfty}frac{1/x}{1}=0$$ Thus, the result is $$lim_{x oinfty} x^{1/x}=e^0=1$$
2. Use the squeeze theorem to show that $$lim_{n oinfty} {n!over n^n}=0$$
Solution:
$$0<frac{n!}{n^n}=frac{1}{n}cdotfrac{2}{n}cdotcdotscdotscdotfrac{n}{n} < frac{1}{n} o0 (n oinfty)$$ According to the squeeze theorem, $$lim_{n oinfty} {n!over n^n}=0$$
3. Determine whether $${sqrt{n+47}-sqrt{n}}_{n=0}^infty$$ converges or diverges. If it converges, compute the limit.
Solution:
$$sqrt{n+47}-sqrt{n}=frac{47}{sqrt{n+47}+sqrt{n}}$$ Hence it is decreasing. On the other hand, $$sqrt{n+47}-sqrt{n}ge0$$ that is, it is bounded below. Thus, it is convergent. And we have $$lim_{x oinfty}(sqrt{n+47}-sqrt{n})=lim_{n oinfty}frac{47}{sqrt{n+47}+sqrt{n}}=0$$ 4. Determine whether $$left{{n^2+1over (n+1)^2} ight}_{n=0}^infty$$ converges or diverges. If it converges, compute the limit.
Solution:
$${(n+1)^2over n^2+1}=1+{2nover n^2+1}$$ which is decreasing. Thus $${n^2+1over (n+1)^2}$$ is increasing. On the other hand, $${n^2+1over (n+1)^2}={n^2+1over n^2+2n+1} < 1$$ which means it is bounded above. Thus it is convergent. And we have $$lim_{n oinfty}{n^2+1over (n+1)^2}=lim_{n oinfty}frac{n^2+1}{n^2+2n+1}=lim_{n oinfty}frac{1+frac{1}{n^2}}{1+frac{2}{n}+frac{1}{n^2}}=1$$
5. Determine whether $$left{{n+47oversqrt{n^2+3n}} ight}_{n=1}^infty$$ converges or diverges. If it converges, compute the limit.
Solution:
$$f^{'}(n)=frac{sqrt{n^2+3n}-(n+47)cdot{1over2}cdot{1oversqrt{n^2+3n}}cdot(2n+3)}{n^2+3n} < 0$$ $$Longleftrightarrow sqrt{n^2+3n}-(n+47)cdot{1over2}cdot{1oversqrt{n^2+3n}}cdot(2n+3) < 0$$ $$Longleftrightarrow n^2+3n < {1over2}cdot(2n^2+97n+141)$$ $$Longleftrightarrow n^2+3n < n^2+48.5n+70.5$$ The last inequality is obvious. Thus it is decreasing. On the other hand, $${n+47oversqrt{n^2+3n}}>0$$ which means it is bounded below. Hence it is convergent. And we have $$lim_{n oinfty}{n+47oversqrt{n^2+3n}}=lim_{n oinfty}{1+frac{47}{n}oversqrt{1+frac{3}{n}}}=1$$
6. Determine whether $$left{{2^nover n!} ight}_{n=0}^infty$$ converges or diverges. If it converges, compute the limit.
Solution:
$${a_{n+1}over a_n}={frac{2^{n+1}}{(n+1)!}overfrac{2^n}{n!}}={2over n+1} < 1$$ when $n > 2$. Thus it is decreasing. On the other hand, $${2^nover n!}>0$$ which means it is bounded below. Thus it is convergent. $$0<{2^nover n!}={2over n}cdot {2over n-1} cdotcdotscdotscdot{2over3}cdot{2over2}cdot{2over1} < ({2over3})^{n-2}cdot2 o0 (n oinfty)$$ According to squeeze theorem we have $$lim_{n oinfty}{2^nover n!}=0$$