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![](https://rs.olm.vn/images/avt/0.png?1311)
Làm theo cách của Trắng nha ,
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\left(Đpcm\right)\)
Ta có: \(\frac{1}{2^2}=\frac{1}{2^2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...................
\(\frac{1}{2019^2}< \frac{1}{2018.2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)
\(=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)
\(=\frac{1}{4}+\frac{2}{4}-\frac{1}{2019}\)
\(=\frac{3}{4}-\frac{1}{2019}\)\(< \frac{3}{4}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)
Điều phải chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}\)
\(\text{Vì}\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2011^2}< \frac{1}{2010.2011}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2011}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2011}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{3}{4}-\frac{1}{2011}< \frac{3}{4}\)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(S=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)
Đặt \(M=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{2019!}\)
\(\Rightarrow M< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\)
\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow M< 1-\frac{1}{2019}=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)
\(\Rightarrow S< 1+\frac{2018}{2019}=\frac{2019}{2019}+\frac{2018}{2019}=\frac{4037}{2019}< 2\)
\(\Rightarrow S< 2\) ( ĐPCM )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này nhiều người đăng lắm,bạn vào câu hỏi tương tự
Đặt B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
Đặt A =\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{3\cdot2}\)
...
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}< 1\)
\(\Rightarrow B< A< 1\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=đã cho.
Ta thấy:
1/2^2<1/1*2(vì 2^2>1*2).
1/3^2<1/2*3(vì 3^2>2*3).
...
1/10^2<1/9*10(vì 10^2>9*10).
=>A<1/1*2+1/2*3+1/3*4+...+1/9*10.
=>A<1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10.
=>A<1-1/10.
=>A<9/10.
Mà 9/10<1.
=>A<1.
Vậy A<1(đpcm).
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu hỏi của Nguyễn Mai Anh - Toán lớp 6 - Học toán với OnlineMath:bạn tham khảo nhé.chỉ khác ở chỗ 45 với 2019 thôi !
Ta thấy :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(.........................\)
\(\frac{1}{2019^2}< \frac{1}{2018.2019}\)
\(\Leftrightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)
\(\Leftrightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}=\frac{2018}{2019}\)
Mà \(0< B< 1\)nên \(B\)không phải là số tự nhiên
~ Hok tốt ~
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2019^2}\)
\(\Rightarrow A=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}=\frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\)
đặt A=1/2^2+....+1/2019^2
vì 1/2^2+....+1/2019^2<1/1.2+1/2.3+....+1/2018.2019
=> A<1/1-1/2+1/2-1/3+.....+1/2018-1/2019
=> A<1-1/2019=2018/2019<3/4.
=> A<3/4.
vậy 1/2^2+....+1/2019^2<3/4