K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

10 tháng 8 2017

k² > k² - 1 = (k-1)(k+1)
⇒ 1/k² < 1/[(k-1).(k+1)] = [1/(k-1) - 1/(k+1)]/2 (*)

Áp dụng (*), ta có:
1/2² + 1/3² + 1/4² + ... + 1/n²
< 1/2² + 1/(2.4) + 1/(3.5) + ... + 1/[(n-1).(n+1)]
= 1/2² + [1/2 - 1/4 + 1/3 - 1/5 + ... + 1/(n-1) - 1/(n+1)]/2
= 1/2² + [1/2 + 1/3 - 1/n - 1/(n+1)]/2
= 2/3 - [1/n + 1/(n+1)]/2 < 2/3

10 tháng 8 2017

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right).n}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{\left(n-1\right)}-\dfrac{1}{n}=1-\dfrac{1}{n}\le1-\dfrac{1}{2}=\dfrac{1}{2}< \dfrac{2}{3}\)

\(\Rightarrow\)đpcm

3 tháng 12 2017

Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)

\(\Rightarrow A< \dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)

\(\Rightarrow2A< \dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(n-1\right)\left(n+1\right)}\)

\(\Rightarrow2A< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n+1}\)

\(\Rightarrow2A< 1\)

\(\Rightarrow A< \dfrac{1}{2}< \dfrac{2}{3}\)

NV
25 tháng 3 2023

\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)

\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)

11 tháng 12 2021

\(S=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{n^2}\right)\\ S=\left(1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)\\ S=n-1-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)< n-1\)

Lại có \(\dfrac{1}{4}+\dfrac{1}{9}+..+\dfrac{1}{n^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)

\(\Rightarrow S>n-1-1=n-2\\ \Rightarrow n-2< S< n-1\\ \Rightarrow S\notin N\)

NV
30 tháng 6 2021

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Do đó:

\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)

19 tháng 8 2017

Đặt :

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{n^2}\)

Ta thấy :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

..........................

\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{\left(n-1\right)n}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Leftrightarrow A< 1-\dfrac{1}{n}< 1\)

\(\Leftrightarrow A< 1\)

Vậy ......

19 tháng 8 2017

A<1 nha banvui

27 tháng 12 2021

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)

\(S=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)

=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)

=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)