Đặt: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2019}{3^{2019}}\)

\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2019}{3^{2018}}\)

\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)

Đặt: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)

\(\Rightarrow2B=1-\frac{1}{3^{2018}}\)

\(\Rightarrow B=\frac{1-\frac{1}{3^{2018}}}{2}\)

Thay vào \(2A\Rightarrow2A=1+\frac{\left(1-\frac{1}{3^{2018}}\right)}{2}-\frac{2019}{3^{2019}}\)

\(=1+\frac{1}{2}-\frac{1}{2.3^{2018}}-\frac{2019}{3^{2019}}< 1+\frac{1}{2}=\frac{3}{2}\)

\(\Rightarrow A< 0,75\left(đpcm\right)\)

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

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}\)

Ta có:

\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

....

\(\frac{1}{2019^2}=\frac{1}{2019.2019}< \frac{1}{2018.2019}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow A< 1-\frac{1}{2019}\)

\(\Rightarrow A< \frac{2018}{2019}\)

đến đây mới thấy mik sai ,xin lỗi

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

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)\(+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{2018}{2019}\)

Mà: \(\frac{3}{4}=\frac{2016}{2688}< \frac{2017}{2688}< \frac{2017}{2019}< \frac{2018}{2019}\)

\(\Rightarrow\frac{3}{4}< \frac{2018}{2019}\)

a/

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(A=2A-A=1-\frac{1}{2^{100}}< 1\)

b/

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)

\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)

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 )

Chứng tỏ A < \(\frac{9}{2}\)

Tham khảo

https://hoc24.vn/hoi-dap/question/814814.html

B=11.2+13.4+15.6+....+12019.2020

⇒2B=21.2+23.4+25.6+....+22019.2020

<1+12.3+13.4+14.5+15.6+....+12018.2019+12019.2020

2B<1+3−22.3+4−33.4+5−44.5+....+2019−20182018.2019+2020−20192019.2020

2B<1+12−13+13−14+...+12019−12020

2B<1+12−12020<1+12

B<34

---------------------

Đặt 22018=a;32019=b;52020=c(a,b,c>0)

A=aa+b+bb+c+cc+a>aa+b+c+ba+b+c+ca+b+c=1

⇒A>1>34>B

\(\hept{\begin{cases}A=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}^{ }\\B=-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\end{cases}}\)

=>\(A-B=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}+\frac{1}{2020}+\frac{7}{2019^2}+\frac{5}{2019^3}+\frac{3}{2019^4}\)

\(=>A-B=\left(-\frac{3}{2019^2}+\frac{7}{2019^2}\right)+\left(-\frac{7}{2019^4}+\frac{3}{2019^4}\right)\)

=>\(A-B=\frac{4}{2019^2}+-\frac{4}{2019^4}\)

=>\(A-B=\frac{2019^2.4}{2019^4}-\frac{4}{2019^4}\)

=>\(A>B\)

cách này mình tự nghĩ 

thank you \(v\text{er}y^{1000000000000}\)much

sao ko có ai giúp mk vậy