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ln n, n^k , a^n, n!, The limit problem of n^n

2022-06-22 04:26:00 【Love 123 haha】

ln ⁡ n < < n k < < a n < < n ! < < n n a n < < b n surface in n → ∞ when , a n far far Small On b n , be lim ⁡ n → ∞ a n b n = 0 \ln n<< n^k<<a^n<<n!<<n^n\\ \\a_n<<b_n Express n\rightarrow\infty when ,a_n Far less than b_n, be \lim\limits_{n\rightarrow \infty}\frac{a_n}{b_n}=0 lnn<<nk<<an<<n!<<nnan​<<bn​ surface in n→∞ when ,an​ far far Small On bn​, be n→∞lim​bn​an​​=0

lim ⁡ n → ∞ ln ⁡ n n k = 0 , k ∈ N + \lim\limits_{n\rightarrow\infty}\frac{\ln n}{n^k}=0,k\in \N_+ n→∞lim​nklnn​=0,k∈N+​
k = 1 when k=1 when k=1 when
ln ⁡ n n = ln ⁡ n 1 n = ln ⁡ n n \frac{\ln n}{n}=\ln n^{\frac{1}{n}}=\ln \sqrt[n]{n} nlnn​=lnnn1​=lnnn​
lim ⁡ n → ∞ n n = 1 , n n ≥ 1 \lim\limits_{n\rightarrow \infty} \sqrt[n]{n} = 1,\sqrt[n]{n}\geq1 n→∞lim​nn​=1,nn​≥1
∀ ε , ∃ N , ∀ n > N , n n < e ε , ∣ ln ⁡ n n ∣ = ln ⁡ n n < ε \forall \varepsilon, \exist N,\forall n>N,\sqrt[n]{n}<e^\varepsilon,|\ln\sqrt[n]{n}|=\ln\sqrt[n]{n}<\varepsilon ∀ε,∃N,∀n>N,nn​<eε,∣lnnn​∣=lnnn​<ε
therefore lim ⁡ n → ∞ ln ⁡ n n = 0 \lim\limits_{n\rightarrow\infty}\frac{\ln n}{n}=0 n→∞lim​nlnn​=0
k > 1 k>1 k>1 when
ln ⁡ n n k = ln ⁡ n n ⋅ 1 n k − 1 \frac{\ln n}{n^k}=\frac{\ln n}{n}\cdot\frac{1}{n^{k-1}} nklnn​=nlnn​⋅nk−11​
therefore lim ⁡ n → ∞ ln ⁡ n n k = 0 \lim\limits_{n\rightarrow\infty}\frac{\ln n}{n^k}=0 n→∞lim​nklnn​=0

lim ⁡ n → ∞ n k a n = 0 , a > 1 \lim\limits_{n\rightarrow\infty}\frac{n^k}{a^n}=0,a>1 n→∞lim​annk​=0,a>1
Make x n = n k a n Make x_n=\frac{n^k}{a^n} Make xn​=annk​, be x n > 0 x_n>0 xn​>0
lim ⁡ n → ∞ x n + 1 x n = 1 a lim ⁡ n → ∞ ( 1 + 1 n ) k = 1 a < 1 \lim\limits_{n\rightarrow\infty}\frac{x_{n+1}}{x_n}=\frac{1}{a}\lim\limits_{n\rightarrow\infty}(1+\frac{1}{n})^k=\frac{1}{a}<1 n→∞lim​xn​xn+1​​=a1​n→∞lim​(1+n1​)k=a1​<1
∃ N , ∀ n > N , x n + 1 x n < 1 , the With { x n } from n > N after open beginning yes single transfer Deliver reduce Of \exist N,\forall n>N,\frac{x_{n+1}}{x_n}<1, therefore \{x_n\} from n>N Then it starts to decrease monotonically ∃N,∀n>N,xn​xn+1​​<1, the With { xn​} from n>N after open beginning yes single transfer Deliver reduce Of
x n > 0 , the With { x n } Yes Next world , the With { x n } n > N closed Convergence , be { x n } closed Convergence x_n>0, therefore \{x_n\} There is a lower bound , therefore \{x_n\}_{n>N} convergence , be \{x_n\} convergence xn​>0, the With { xn​} Yes Next world , the With { xn​}n>N​ closed Convergence , be { xn​} closed Convergence
set up lim ⁡ n → ∞ x n = A set up \lim\limits_{n\rightarrow\infty}x_n=A set up n→∞lim​xn​=A
x n + 1 = x n ⋅ 1 a ( 1 + 1 n ) k x_{n+1}=x_n\cdot \frac{1}{a}(1+\frac{1}{n})^k xn+1​=xn​⋅a1​(1+n1​)k
two edge Same as take extremely limit have to , A = A a , the With A = 0 Take the limit on both sides and get ,A=\frac{A}{a}, therefore A=0 two edge Same as take extremely limit have to ,A=aA​, the With A=0
the With lim ⁡ n → ∞ n k a n = 0 , a > 1 therefore \lim\limits_{n\rightarrow\infty}\frac{n^k}{a^n}=0,a>1 the With n→∞lim​annk​=0,a>1

lim ⁡ n → ∞ a n n ! = 0 , a > 0 \lim\limits_{n\rightarrow\infty}\frac{a^n}{n!}=0,a>0 n→∞lim​n!an​=0,a>0
0 < a < 1 when 0<a<1 when 0<a<1 when
0 < a n ⋅ a n − 1 ⋅ . . . ⋅ a 1 < a n 0<\frac{a}{n}\cdot\frac{a}{n-1}\cdot...\cdot\frac{a}{1}<a^n 0<na​⋅n−1a​⋅...⋅1a​<an
lim ⁡ n → ∞ a n = 0 \lim\limits_{n\rightarrow\infty}a^n=0 n→∞lim​an=0
the With lim ⁡ n → ∞ a n n ! = 0 therefore \lim\limits_{n\rightarrow\infty}\frac{a^n}{n!}=0 the With n→∞lim​n!an​=0
a > 1 when a>1 when a>1 when
0 < a n ⋅ a n − 1 ⋅ . . . ⋅ a 1 < a n ⋅ a a ⋅ a a − 1 ⋅ . . . ⋅ a 1 0<\frac{a}{n}\cdot\frac{a}{n-1}\cdot...\cdot\frac{a}{1}<\frac{a}{n}\cdot\frac{a}{a}\cdot\frac{a}{a-1}\cdot...\cdot\frac{a}{1} 0<na​⋅n−1a​⋅...⋅1a​<na​⋅aa​⋅a−1a​⋅...⋅1a​
lim ⁡ n → ∞ a n ⋅ a a ⋅ a a − 1 ⋅ . . . ⋅ a 1 = 0 \lim\limits_{n\rightarrow\infty}\frac{a}{n}\cdot\frac{a}{a}\cdot\frac{a}{a-1}\cdot...\cdot\frac{a}{1}=0 n→∞lim​na​⋅aa​⋅a−1a​⋅...⋅1a​=0
the With lim ⁡ n → ∞ a n n ! = 0 therefore \lim\limits_{n\rightarrow\infty}\frac{a^n}{n!}=0 the With n→∞lim​n!an​=0
a = 1 when a=1 when a=1 when
0 < 1 n ⋅ 1 n − 1 ⋅ . . . ⋅ 1 1 < 1 n 0<\frac{1}{n}\cdot\frac{1}{n-1}\cdot...\cdot\frac{1}{1}<\frac{1}{n} 0<n1​⋅n−11​⋅...⋅11​<n1​
lim ⁡ n → ∞ 1 n = 0 \lim\limits_{n\rightarrow\infty}\frac{1}{n}=0 n→∞lim​n1​=0
the With lim ⁡ n → ∞ a n n ! = 0 therefore \lim\limits_{n\rightarrow\infty}\frac{a^n}{n!}=0 the With n→∞lim​n!an​=0

lim ⁡ n → ∞ n ! n n = 0 \lim\limits_{n\rightarrow\infty}\frac{n!}{n^n}=0 n→∞lim​nnn!​=0
Make x n = n ! n n Make x_n=\frac{n!}{n^n} Make xn​=nnn!​, be x n > 0 x_n>0 xn​>0
x n x n + 1 = ( 1 + 1 n ) n > 1 \frac{x_{n}}{x_{n+1}}=(1+\frac{1}{n})^{n}>1 xn+1​xn​​=(1+n1​)n>1
the With { x n } yes single transfer Deliver reduce Count Column therefore \{x_n\} Is a monotonically decreasing sequence the With { xn​} yes single transfer Deliver reduce Count Column
x n > 0 , the With { x n } Yes Next world , the With { x n } closed Convergence x_n>0, therefore \{x_n\} There is a lower bound , therefore \{x_n\} convergence xn​>0, the With { xn​} Yes Next world , the With { xn​} closed Convergence
set up lim ⁡ n → ∞ x n = A set up \lim\limits_{n\rightarrow\infty}x_n=A set up n→∞lim​xn​=A
x n ⋅ ( 1 + 1 n ) n = x n + 1 x_n\cdot (1+\frac{1}{n})^{n}=x_{n+1} xn​⋅(1+n1​)n=xn+1​
two edge Same as take extremely limit have to , e A = A , the With A = 0 Take the limit on both sides and get ,eA=A, therefore A=0 two edge Same as take extremely limit have to ,eA=A, the With A=0
the With lim ⁡ n → ∞ n ! n n = 0 therefore \lim\limits_{n\rightarrow\infty}\frac{n!}{n^n}=0 the With n→∞lim​nnn!​=0

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