Bài 104679

Question

a. Chứng minh rằng:

$\ln k \leq \int\limits_{k}^{k+1} \ln x dx\leq \ln (k+1), \forall k>0$
b. Chứng minh rằng:

$(n-1)!\leq n^{n}.e^{-n}.e\leq n!,\forall n\in N$ \

$\left\{ \begin{array}{l} \end{array} \right.\left. 0,1 \right \}$
c.Tính :

$\mathop {\lim }\limits_{n \to +\infty }\frac{\sqrt {n!}}{n}$
(Biết:

$\mathop {\lim }\limits_{n \to +\infty } \sqrt[n]{a}= \mathop {\lim }\limits_{n \to +\infty } \sqrt[n]{n} =1,a>0$ )

score 0 · Answer 1

a.Với:

$\forall x \in [k,k+1]\Rightarrow \ln k \leq\ln x \leq \ln (k+1)$

$\Rightarrow \int\limits_{k}^{k+1} \ln k dx \leq \int\limits_{k}^{k+1} \ln x dx\leq\int\limits_{k}^{k+1} \ln (k+1) dx$

$\Rightarrow \ln k \leq \int\limits_{k}^{k+1} \ln x dx\leq \ln (k+1)$

$\Rightarrow$ (ĐPCM)

b.Theo câu a:

$\ln 1+\ln 2+...+\ln (n-1)\leq \int\limits_{1}^{2}\ln x dx+ \int\limits_{2}^{3}\ln x dx+...+ \int\limits_{n-1}^{n}\ln x dx$

$\leq \ln 2+\ln 3+...+\ln n, \forall n\in N$ \

$\left\{ \begin{array}{l} \end{array} \right.\left. 0,1 \right \}$

$\Rightarrow \ln [(n-1)!]\leq \int\limits_{1}^{n}\ln x dx\leq \ln[n!]$

$\Rightarrow \ln [(n-1)!]\leq [x\ln \frac{x}{e}]_{1}^{n}\leq \ln[n!]$

$\Rightarrow \ln [(n-1)!]\leq n\ln \frac{n}{e}-\ln \frac{1}{e}\leq \ln[n!]$

$\Rightarrow \ln [(n-1)!]\leq \ln[(\frac{n}{e})^{n}.e]\leq \ln[n!]$

$\Rightarrow (n-1)!\leq (\frac{n}{e})^{n}.e\leq n!$

$\Rightarrow(n-1)!\leq n^{n}.e^{-n}.e\leq n!$

$\Rightarrow$ (ĐPCM)

c.Theo b:

$(n-1)!\leq n^{n}.e^{-n}.e\leq n!$

$\Rightarrow \frac{1}{n}n!\leq n^{n}.e^{-n}.e\leq n!$

$\Rightarrow \frac{1}{\sqrt[n]{n}}.\sqrt[n]{n!}\leq n.e^{-1}.\sqrt[n]{e}\leq \sqrt[n]{n!}$

$\Rightarrow \frac{\sqrt[n]{e}}{e}\leq \frac{\sqrt[n]{n!}}{n}\leq \frac{\sqrt[n]{n}.\sqrt[n]{e}}{e}$

Do:

$\mathop {\lim }\limits_{n \to +\infty } \frac{\sqrt[n]{e}}{e}=\mathop {\lim }\limits_{n \to +\infty }\frac{\sqrt[n]{n}.\sqrt[n]{e}}{e}=\frac{1}{e}$

$\Rightarrow \mathop {\lim }\limits_{n \to +\infty }\frac{\sqrt[n]{n!}}{n}=\frac{1}{e}$

1 Đáp án