【数学分析笔记】第2章第2节数列极限(2)
2. 数列极限
2.2 数列极限
【例2.2.3】证明 lim n → ∞ a n = 1 ( a > 1 ) \lim\limits_{n\to\infty}\sqrt[n]{a}=1(a>1) n→∞limna=1(a>1)
【证】对任意给定的 ε > 0 \varepsilon >0 ε>0,令 a n − 1 = y n > 0 ( a > 1 ) \sqrt[n]{a}-1=y_{n}>0(a>1) na−1=yn>0(a>1)
a n = 1 + y n \sqrt[n]{a}=1+y_{n} na=1+yn,两边同时取 n n n次方得
a = ( 1 + y n ) 2 = 1 + C n 1 y n + C n 2 y n 2 + . . . + C n n y n n > 1 + n y n a=(1+y_{n})^{2}=1+C_{n}^{1}y_{n}+C_{n}^{2}y_{n}^{2}+...+C_{n}^{n}y_{n}^{n}>1+ny_{n} a=(1+yn)2=1+Cn1yn+Cn2yn2+...+Cnnynn>1+nyn,则 y n < a − 1 n y_{n}<\frac{a-1}{n} yn<na−1
若要 ∣ a n − 1 ∣ = ∣ y n ∣ = a n − 1 < a − 1 n < ε |\sqrt[n]{a}-1|=|y_{n}|=\sqrt[n]{a}-1<\frac{a-1}{n}<\varepsilon ∣na−1∣=∣yn∣=na−1<na−1<ε即 a − 1 n < ε \frac{a-1}{n}<\varepsilon na−1<ε
则 n > a − 1 ε n>\frac{a-1}{\varepsilon} n>εa−1,由于 a > 1 , ε > 0 a>1,\varepsilon>0 a>1,ε>0,所以 a − 1 ε > 0 \frac{a-1}{\varepsilon}>0 εa−1>0(小于一个大于0的数,就能说明 ε \varepsilon ε, a − 1 ε \frac{a-1}{\varepsilon} εa−1若在区间 ( 0 , 1 ) (0,1) (0,1)内,取整后有可能为0,还得加个1)
所以取 N = [ a − 1 ε ] + 1 N=[\frac{a-1}{\varepsilon}]+1 N=[εa−1]+1
故当 n > N n>N n>N时, ∣ a n − 1 ∣ = a n − 1 < ε |\sqrt[n]{a}-1|=\sqrt[n]{a}-1<\varepsilon ∣na−1∣=na−1<ε
所以 lim n → ∞ a n = 1 ( a > 1 ) \lim\limits_{n\to\infty}\sqrt[n]{a}=1(a>1) n→∞limna=1(a>1)
【例2.2.4】证明 lim n → ∞ n n = 1 \lim\limits_{n\to\infty}\sqrt[n]{n}=1 n→∞limnn=1
【证】对任意给定的 ε > 0 \varepsilon>0 ε>0,
要证 ∣ n n − 1 ∣ = n n − 1 < ε |\sqrt[n]{n}-1|=\sqrt[n]{n}-1<\varepsilon ∣nn−1∣=nn−1<ε(数列的下标 n n n是从1开始计算的,也就是 n ≥ 1 n\ge 1 n≥1,所以 n n ≥ 1 \sqrt[n]{n}\ge 1 nn≥1,即 ∣ n n − 1 ∣ = n n − 1 |\sqrt[n]{n}-1|=\sqrt[n]{n}-1 ∣nn−1∣=nn−1)
令 n n − 1 = y n > 0 ( n = 2 , 3 , . . . ) \sqrt[n]{n}-1=y_{n}>0(n=2,3,...) nn−1=yn>0(n=2,3,...)( n n n从2开始 y n > 0 y_{n}>0 yn>0)
n n = y n + 1 ( n = 2 , 3 , . . . ) \sqrt[n]{n}=y_{n}+1(n=2,3,...) nn=yn+1(n=2,3,...)
即 n = 1 + C n 1 y n + C n 2 y n 2 + . . . + C n n y n n > 1 + n ( n − 1 ) 2 y n 2 n=1+C_{n}^{1}y_{n}+C_{n}^{2}y_{n}^{2}+...+C_{n}^{n}y_{n}^{n}>1+\frac{n(n-1)}{2}y_{n}^{2} n=1+Cn1yn+Cn2yn2+...+Cnnynn>1+2n(n−1)yn2
即 y n 2 < 2 n y_{n}^{2}<\frac{2}{n} yn2<n2所以 y n < 2 n y_{n}<\sqrt{\frac{2}{n}} yn<n2(不等式最左侧有 n n n,所以右侧放缩不能放出含 n n n的项)
若要 y n < 2 n < ε y_{n}<\sqrt{\frac{2}{n}}<\varepsilon yn<n2<ε
只要 n > 2 ε 2 n>\frac{2}{\varepsilon ^{2}} n>ε22
取 N = [ 2 ε 2 ] N=[\frac{2}{\varepsilon ^{2}}] N=[ε22]
当 n > N n>N n>N时, ∣ n n − 1 ∣ = n n − 1 < ε |\sqrt[n]{n}-1|=\sqrt[n]{n}-1<\varepsilon ∣nn−1∣=nn−1<ε
即 lim n → ∞ n n = 1 \lim\limits_{n\to\infty}\sqrt[n]{n}=1 n→∞limnn=1
【拓展】证明 lim n → ∞ n 2 n = 1 \lim\limits_{n\to\infty}\sqrt[n]{n^{2}}=1 n→∞limnn2=1
【证(我自己写的,请数院大神批评指正)】对任意给定的 ε > 0 \varepsilon>0 ε>0,
要证 ∣ n 2 n − 1 ∣ = n 2 n − 1 < ε |\sqrt[n]{n^{2}}-1|=\sqrt[n]{n^{2}}-1<\varepsilon ∣nn2−1∣=nn2−1<ε
令 n 2 n − 1 = y n > 0 ( n = 2 , 3 , . . . ) \sqrt[n]{n^{2}}-1=y_{n}>0(n=2,3,...) nn2−1=yn>0(n=2,3,...)( n n n从2开始 y n > 0 y_{n}>0 yn>0)
n 2 n = y n + 1 ( n = 2 , 3 , . . . ) \sqrt[n]{n^{2}}=y_{n}+1(n=2,3,...) nn2=yn+1(n=2,3,...)
即 n 2 = 1 + C n 1 y n + C n 2 y n 2 + . . . + C n n y n n > 1 + n ! 3 ! ( n − 3 ) ! y n 3 = 1 + n ( n − 1 ) ( n − 2 ) 3 × 2 × 1 y n 3 = 1 + n ( n − 1 ) ( n − 2 ) 6 y n 3 n^{2}=1+C_{n}^{1}y_{n}+C_{n}^{2}y_{n}^{2}+...+C_{n}^{n}y_{n}^{n}>1+\frac{n!}{3!(n-3)!}y_{n}^{3}=1+\frac{n(n-1)(n-2)}{3\times 2\times 1}y_{n}^{3}=1+\frac{n(n-1)(n-2)}{6}y_{n}^{3} n2=1+Cn1yn+Cn2yn2+...+Cnnynn>1+3!(n−3)!n!yn3=1+3×2×1n(n−1)(n−2)yn3=1+6n(n−1)(n−2)yn3
即 n 2 − 1 > n ( n − 1 ) ( n − 2 ) 6 y n 3 n^{2}-1>\frac{n(n-1)(n-2)}{6}y_{n}^{3} n2−1>6n(n−1)(n−2)yn3
亦即 ( n + 1 ) ( n − 1 ) > n ( n − 1 ) ( n − 2 ) 6 y n 3 (n+1)(n-1)>\frac{n(n-1)(n-2)}{6}y_{n}^{3} (n+1)(n−1)>6n(n−1)(n−2)yn3
所以 6 ( n + 1 ) ( n − 1 ) ( n − 2 ) > y n 3 \frac{6(n+1)}{(n-1)(n-2)}>y_{n}^{3} (n−1)(n−2)6(n+1)>yn3
即 y n 3 < 6 ( n + 1 ) ( n − 1 ) ( n − 2 ) y_{n}^{3}<\frac{6(n+1)}{(n-1)(n-2)} yn3<(n−1)(n−2)6(n+1)所以 y n < 6 ( n + 1 ) ( n − 1 ) ( n − 2 ) 3 < 6 ( n − 1 ) ( n − 1 ) ( n − 2 ) 3 = 6 n − 2 3 y_{n}<\sqrt[3]{\frac{6(n+1)}{(n-1)(n-2)}}<\sqrt[3]{\frac{6(n-1)}{(n-1)(n-2)}}=\sqrt[3]{\frac{6}{n-2}} yn<3(n−1)(n−2)6(n+1)<3(n−1)(n−2)6(n−1)=3n−26(不等式最左侧有 n n n,所以右侧放缩不能放出含 n n n的项)
若要 y n < 6 n − 2 3 < ε y_{n}<\sqrt[3]{\frac{6}{n-2}}<\varepsilon yn<3n−26<ε
只要 n − 2 > 6 ε 3 n-2>\frac{6}{\varepsilon ^{3}} n−2>ε36
即 n > 6 ε 3 + 2 n>\frac{6}{\varepsilon ^{3}}+2 n>ε36+2
取 N = [ 6 ε 3 + 2 ] N=[\frac{6}{\varepsilon ^{3}}+2] N=[ε36+2](这时候肯定满足了 N N N是正整数)
当 n > N n>N n>N时, ∣ n 2 n − 1 ∣ = n 2 n − 1 < ε |\sqrt[n]{n^{2}}-1|=\sqrt[n]{n^{2}}-1<\varepsilon ∣nn2−1∣=nn2−1<ε
即 lim n → ∞ n 2 n = 1 \lim\limits_{n\to\infty}\sqrt[n]{n^{2}}=1 n→∞limnn2=1
…
以此类推, lim n → ∞ n k n = 1 , k ∈ N + \lim\limits_{n\to\infty}\sqrt[n]{n^{k}}=1,k\in\mathbb{N}^{+} n→∞limnnk=1,k∈N+
【例2.2.5】证明 lim n → ∞ n 2 + 1 2 n 2 − 7 n = 1 2 \lim\limits_{n\to\infty}\frac{n^{2}+1}{2n^{2}-7n}=\frac{1}{2} n→∞lim2n2−7nn2+1=21
【证】对任意给定的 ε > 0 \varepsilon>0 ε>0
要使得 ∣ n 2 + 1 2 n 2 − 7 n − 1 2 ∣ = ∣ 2 n 2 + 2 − 2 n 2 + 7 n 2 ( 2 n 2 − 7 n ) ∣ = ∣ 7 n + 2 2 ( 2 n 2 − 7 n ) ∣ |\frac{n^{2}+1}{2n^{2}-7n}-\frac{1}{2}|=|\frac{2n^{2}+2-2n^{2}+7n}{2(2n^{2}-7n)}|=|\frac{7n+2}{2(2n^{2}-7n)}| ∣2n2−7nn2+1−21∣=∣2(2n2−7n)2n2+2−2n2+7n∣=∣2(2n2−7n)7n+2∣
当 n > 3 n>3 n>3(找 N N N比3大就行,数列的有限项不影响数列极限取值)时,
则 ∣ n 2 + 1 2 n 2 − 7 n − 1 2 ∣ = 7 n + 2 2 ( 2 n 2 − 7 n ) |\frac{n^{2}+1}{2n^{2}-7n}-\frac{1}{2}|=\frac{7n+2}{2(2n^{2}-7n)} ∣2n2−7nn2+1−21∣=2(2n2−7n)7n+2
当 n > 3 n>3 n>3的时候, 7 n + 2 < 7 n + n = 8 n 7n+2<7n+n=8n 7n+2<7n+n=8n
为了能够放缩出更简单的形式,我们要想尽办法找到比 2 ( 2 n 2 − 7 n ) = 2 n ( 2 n − 7 ) 2(2n^{2}-7n)=2n(2n-7) 2(2n2−7n)=2n(2n−7)分母小的,分母小,整个分式越大, 2 n ( 2 n − 7 ) 2n(2n-7) 2n(2n−7)不可能大于 4 n 2 4n^{2} 4n2,因为那样会出现 − 7 = 0 -7=0 −7=0的矛盾情况,于是尝试找 2 n ( 2 n − 7 ) > 2 n ⋅ n 2n(2n-7)>2n\cdot n 2n(2n−7)>2n⋅n的情况,即 2 n − 7 > n 2n-7>n 2n−7>n即 n > 7 n>7 n>7(前7项都是有限项,不耽误后面无限项趋近于极限值)的时候,有:
2 ( 2 n 2 − 7 n ) = 2 n ( 2 n − 7 ) = − 2 n 2(2n^{2}-7n)=2n(2n-7)=-2n 2(2n2−7n)=2n(2n−7)=−2n
若要 7 n + 2 2 ( 2 n 2 − 7 n ) < 8 n 2 n 2 = 4 n < ε \frac{7n+2}{2(2n^{2}-7n)}<\frac{8n}{2n^{2}}=\frac{4}{n}<\varepsilon 2(2n2−7n)7n+2<2n28n=n4<ε,还要保证 n > 7 n>7 n>7
即取 N = max { [ 4 ε ] , 7 } N=\max\{[\frac{4}{\varepsilon}],7\} N=max{[ε4],7}(视频课中写的是 6 6 6,也可以,保证 2 n ( 2 n − 7 ) ≥ 2 n ⋅ n 2n(2n-7)\ge2n\cdot n 2n(2n−7)≥2n⋅n,则 7 n + 2 2 ( 2 n 2 − 7 n ) ≤ 8 n 2 n 2 = 4 n < ε \frac{7n+2}{2(2n^{2}-7n)}\le\frac{8n}{2n^{2}}=\frac{4}{n}<\varepsilon 2(2n2−7n)7n+2≤2n28n=n4<ε,最后是小于 ε \varepsilon ε且能将不等式不等号方向顺下来就行)
当 n > N n>N n>N时, ∣ n 2 + 1 2 n 2 − 7 n − 1 2 ∣ < 4 n < ε |\frac{n^{2}+1}{2n^{2}-7n}-\frac{1}{2}|<\frac{4}{n}<\varepsilon ∣2n2−7nn2+1−21∣<n4<ε
所以 lim n → ∞ n 2 + 1 2 n 2 − 7 n = 1 2 \lim\limits_{n\to\infty}\frac{n^{2}+1}{2n^{2}-7n}=\frac{1}{2} n→∞lim2n2−7nn2+1=21
【例2.2.6】设 lim n → ∞ a n = a \lim\limits_{n\to\infty}a_{n}=a n→∞liman=a,证明 lim n → ∞ a 1 + a 2 + . . . + a n n = a \lim\limits_{n\to\infty}\frac{a_{1}+a_{2}+...+a_{n}}{n}=a n→∞limna1+a2+...+an=a
【证】(1)设 a = 0 a=0 a=0对任意给定的 ε > 0 \varepsilon>0 ε>0,存在 N 1 N_{1} N1,当 n > N 1 n>N_{1} n>N1时, ∣ a n ∣ < ε 2 |a_{n}|<\frac{\varepsilon}{2} ∣an∣<2ε(任意给定,只要保证大于0就行,所以取 ε = ε 2 > 0 \varepsilon=\frac{\varepsilon}{2}>0 ε=2ε>0)
a 1 + a 2 + . . . + a n n = a 1 + a 2 + . . . + a N 1 n + a N 1 + 1 + a N 1 + 2 + . . . + a n n \frac{a_{1}+a_{2}+...+a_{n}}{n}=\frac{a_{1}+a_{2}+...+a_{N_{1}}}{n}+\frac{a_{N_{1}+1}+a_{N_{1}+2}+...+a_{n}}{n} na1+a2+...+an=na1+a2+...+aN1+naN1+1+aN1+2+...+an
由于当 n > N 1 n>N_{1} n>N1时, ∣ a n ∣ < ε 2 |a_{n}|<\frac{\varepsilon}{2} ∣an∣<2ε(讨论 a a a为0也是为了这里方便一些,更是为了后续证明 a ≠ 0 a\ne 0 a=0的情况),且 N 1 + 1 , N 1 + 2 , . . . , a n > N 1 N_{1}+1,N_{1}+2,...,a_{n}>N_{1} N1+1,N1+2,...,an>N1,所以 ∣ a N 1 + 1 + a N 1 + 2 + . . . + a n n ∣ ≤ ∣ a N 1 + 1 ∣ + ∣ a N 1 + 2 ∣ + . . . + ∣ a n ∣ n < ( n − ( N 1 + 1 ) + 1 ) ⋅ ε 2 n = ( n − N 1 ) ⋅ ε 2 n < n ⋅ ε 2 n = ε 2 |\frac{a_{N_{1}+1}+a_{N_{1}+2}+...+a_{n}}{n}|\le\frac{|a_{N_{1}+1}|+|a_{N_{1}+2}|+...+|a_{n}|}{n}<\frac{(n-(N_{1}+1)+1)\cdot\frac{\varepsilon}{2}}{n}=\frac{(n-N_{1})\cdot\frac{\varepsilon}{2}}{n}<\frac{n\cdot\frac{\varepsilon}{2}}{n}=\frac{\varepsilon}{2} ∣naN1+1+aN1+2+...+an∣≤n∣aN1+1∣+∣aN1+2∣+...+∣an∣<n(n−(N1+1)+1)⋅2ε=n(n−N1)⋅2ε<nn⋅2ε=2ε
由于 a 1 + a 2 + . . . + a N 1 a_{1}+a_{2}+...+a_{N_{1}} a1+a2+...+aN1是确定的有限项的和,则取 N > N 1 N>N_{1} N>N1,使得 n > N > N 1 n>N>N_{1} n>N>N1时,有 ∣ a 1 + a 2 + . . . + a N 1 n ∣ < ε 2 |\frac{a_{1}+a_{2}+...+a_{N_{1}}}{n}|<\frac{\varepsilon}{2} ∣na1+a2+...+aN1∣<2ε(这里陈纪修老师应该是默认我们证明过了 lim n → ∞ c n = 0 , c \lim\limits_{n\to\infty}\frac{c}{n}=0,c n→∞limnc=0,c为常数,取 N > N 1 N>N_{1} N>N1是为了保证 a 1 + a 2 + . . . + a N 1 a_{1}+a_{2}+...+a_{N_{1}} a1+a2+...+aN1是有限项,这样有限项的和就是一个常数,记 a 1 + a 2 + . . . + a N 1 = c a_{1}+a_{2}+...+a_{N_{1}}=c a1+a2+...+aN1=c,对于任意给定的 ε > 0 \varepsilon >0 ε>0,则 ∣ c n − 0 ∣ = ∣ c n ∣ = ∣ c ∣ n < ε 2 |\frac{c}{n}-0|=|\frac{c}{n}|=\frac{|c|}{n}<\frac{\varepsilon}{2} ∣nc−0∣=∣nc∣=n∣c∣<2ε,取 N = [ 2 ∣ c ∣ ε ] + 1 N=[\frac{2|c|}{\varepsilon}]+1 N=[ε2∣c∣]+1,当 n > N n>N n>N时,有 ∣ c n − 0 ∣ = ∣ c n ∣ < ε 2 |\frac{c}{n}-0|=|\frac{c}{n}|<\frac{\varepsilon}{2} ∣nc−0∣=∣nc∣<2ε,本题这里是相当于是知道了 lim n → ∞ c n = 0 \lim\limits_{n\to\infty}\frac{c}{n}=0 n→∞limnc=0推出 ∣ c n ∣ < ε 2 |\frac{c}{n}|<\frac{\varepsilon}{2} ∣nc∣<2ε,这里想了好久,要不然这篇博客也不会这么久不发出去)
所以 ∣ a 1 + a 2 + . . . + a n n ∣ = ∣ a 1 + a 2 + . . . + a N 1 n ∣ + ∣ a N 1 + 1 + a N 1 + 2 + . . . + a n n ∣ < ε 2 + ε 2 = ε |\frac{a_{1}+a_{2}+...+a_{n}}{n}|=|\frac{a_{1}+a_{2}+...+a_{N_{1}}}{n}|+|\frac{a_{N_{1}+1}+a_{N_{1}+2}+...+a_{n}}{n}|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon ∣na1+a2+...+an∣=∣na1+a2+...+aN1∣+∣naN1+1+aN1+2+...+an∣<2ε+2ε=ε
所以 lim n → ∞ a 1 + a 2 + . . . + a n n = a = 0 \lim\limits_{n\to\infty}\frac{a_{1}+a_{2}+...+a_{n}}{n}=a=0 n→∞limna1+a2+...+an=a=0
(2)当 a ≠ 0 a\ne 0 a=0时,由于 lim n → ∞ a n = a \lim\limits_{n\to\infty}a_{n}=a n→∞liman=a,则 { a n − a } \{a_{n}-a\} {an−a}的极限是无穷小量.(上一节课无穷小量的定义)
lim n → ∞ ( a 1 + a 2 + . . . + a n n − a ) = lim n → ∞ a 1 − a + a 2 − a + . . . + a n − a n \lim\limits_{n\to\infty}(\frac{a_{1}+a_{2}+...+a_{n}}{n}-a)=\lim\limits_{n\to\infty}\frac{a_{1}-a+a_{2}-a+...+a_{n}-a}{n} n→∞lim(na1+a2+...+an−a)=n→∞limna1−a+a2−a+...+an−a
由于(1)中已经证明的结论,由于 { a n − a } \{a_{n}-a\} {an−a}的极限是无穷小量即 lim n → ∞ ( a n − a ) = 0 \lim\limits_{n\to\infty}(a_{n}-a)=0 n→∞lim(an−a)=0,所以 lim n → ∞ ( a 1 + a 2 + . . . + a n n − a ) = lim n → ∞ a 1 − a + a 2 − a + . . . + a n − a n = 0 ⇔ lim n → ∞ a 1 + a 2 + . . . + a n n = a \lim\limits_{n\to\infty}(\frac{a_{1}+a_{2}+...+a_{n}}{n}-a)=\lim\limits_{n\to\infty}\frac{a_{1}-a+a_{2}-a+...+a_{n}-a}{n}=0\Leftrightarrow\lim\limits_{n\to\infty}\frac{a_{1}+a_{2}+...+a_{n}}{n}=a n→∞lim(na1+a2+...+an−a)=n→∞limna1−a+a2−a+...+an−a=0⇔n→∞limna1+a2+...+an=a
【注】用到了上节课的知识:若 lim n → ∞ x n = a ⇔ { x n − a } \lim\limits_{n\to \infty}x_{n}=a\Leftrightarrow\{x_{n}-a\} n→∞limxn=a⇔{xn−a}是无穷小量。
2.2.5 数列极限的性质
记号: ∀ \forall ∀ 对于每一个,任意的
∃ \exists ∃ 存在,可以找到
lim n → ∞ x n = a \lim\limits_{n\to\infty}x_{n}=a n→∞limxn=a的定义用记号写为: ∀ ε > 0 , ∃ N , ∀ n > N : ∣ x n − a ∣ < ε \forall\varepsilon>0,\exists N,\forall n>N: |x_{n}-a|<\varepsilon ∀ε>0,∃N,∀n>N:∣xn−a∣<ε
- 唯一性:若 lim n → ∞ x n = a , lim n → ∞ x n = b \lim\limits_{n\to\infty}x_{n}=a,\lim\limits_{n\to\infty}x_{n}=b n→∞limxn=a,n→∞limxn=b,则 a = b a=b a=b
【证】 ∀ ε > 0 , ∃ N 1 , ∀ n > N 1 : ∣ x n − a ∣ < ε 2 \forall\varepsilon>0,\exists N_{1},\forall n>N_{1}:|x_{n}-a|<\frac{\varepsilon}{2} ∀ε>0,∃N1,∀n>N1:∣xn−a∣<2ε(取得巧妙的 ε 2 \frac{\varepsilon}{2} 2ε)
∀ ε > 0 , ∃ N 2 , ∀ n > N 2 : ∣ x n − a ∣ < ε 2 \forall\varepsilon>0,\exists N_{2},\forall n>N_{2}:|x_{n}-a|<\frac{\varepsilon}{2} ∀ε>0,∃N2,∀n>N2:∣xn−a∣<2ε
取 N = max { N 1 , N 2 } , ∀ n > N : N=\max\{N_{1},N_{2}\},\forall n>N: N=max{N1,N2},∀n>N:
由三角不等式 ∣ ∣ a ∣ − ∣ b ∣ ∣ ≤ ∣ a + b ∣ ≤ ∣ a ∣ + ∣ b ∣ ||a|-|b||\le |a+b| \le |a|+|b| ∣∣a∣−∣b∣∣≤∣a+b∣≤∣a∣+∣b∣
∣ x n − a ∣ + ∣ x n − b ∣ = ∣ a − x n ∣ + ∣ x n − b ∣ ≥ ∣ a − x n + x n − b ∣ = ∣ a − b ∣ |x_{n}-a|+|x_{n}-b|=|a-x_{n}|+|x_{n}-b|\ge|a-x_{n}+x_{n}-b|=|a-b| ∣xn−a∣+∣xn−b∣=∣a−xn∣+∣xn−b∣≥∣a−xn+xn−b∣=∣a−b∣
则 ∣ a − b ∣ ≤ ∣ x n − a ∣ + ∣ x n − b ∣ < ε 2 + ε 2 = ε |a-b|\le|x_{n}-a|+|x_{n}-b|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon ∣a−b∣≤∣xn−a∣+∣xn−b∣<2ε+2ε=ε
由 ε \varepsilon ε的任意性可知, a = b a=b a=b - 有界性:有数列 { x n } \{x_{n}\} {xn},若 ∃ M ∈ R , ∀ n ∈ N + \exists M\in\mathbb{R},\forall n\in\mathbb{N}^{+} ∃M∈R,∀n∈N+成立 x n ≤ M x_{n}\le M xn≤M,则 M M M是 { x n } \{x_{n}\} {xn}的一个上界,或称 { x n } \{x_{n}\} {xn}有上界;若 ∃ m ∈ R , ∀ n ∈ N + \exists m\in\mathbb{R},\forall n\in\mathbb{N}^{+} ∃m∈R,∀n∈N+成立 x n ≥ m x_{n}\ge m xn≥m,则 m m m是 { x n } \{x_{n}\} {xn}的一个下界,或称 { x n } \{x_{n}\} {xn}有下界; { x n } \{x_{n}\} {xn}既有上界又有下界,则称 { x n } \{x_{n}\} {xn}有界;有界的另一定义: ∃ X ∈ R + , ∀ n ∈ N + \exists X\in\mathbb{R}^{+},\forall n\in\mathbb{N}^{+} ∃X∈R+,∀n∈N+成立,则 ∣ x n ∣ ≤ X |x_{n}|\le X ∣xn∣≤X.
【定理2.2.2】收敛数列一定有界。
【证】若 { x n } \{x_{n}\} {xn}收敛于 a a a, ∀ ε > 0 , ∃ N , ∀ n > N : ∣ x n − a ∣ < ε \forall\varepsilon>0,\exists N,\forall n>N: |x_{n}-a|<\varepsilon ∀ε>0,∃N,∀n>N:∣xn−a∣<ε,取 ε = 1 \varepsilon=1 ε=1,存在 N , ∀ n > N : ∣ x n − a ∣ < 1 ⇔ a − 1 < x n < a + 1 N,\forall n>N:|x_{n}-a|<1\Leftrightarrow a-1<x_{n}<a+1 N,∀n>N:∣xn−a∣<1⇔a−1<xn<a+1
取 M = max { x 1 , x 2 , . . . , x n , a + 1 } , m = min { x 1 , x 2 , . . . , x n , a − 1 } M=\max\{x_{1},x_{2},...,x_{n},a+1\},m=\min\{x_{1},x_{2},...,x_{n},a-1\} M=max{x1,x2,...,xn,a+1},m=min{x1,x2,...,xn,a−1}(有限个数,找到数列中的最大值项和 a + 1 a+1 a+1取最大值,找到数列中的最小值与 a − 1 a-1 a−1取最小值,那么这个最大值肯定是 { x n } \{x_{n}\} {xn}的一个上界,最小值肯定是 { x n } \{x_{n}\} {xn}的一个下界)
∀ n ∈ N + , m ≤ x n ≤ M \forall n\in\mathbb{N}^{+},m\le x_{n}\le M ∀n∈N+,m≤xn≤M
即收敛数列一定有界。