Ta có:
B=1-1/2²-1/3²-...-1/2004²
=1-(1/2²+1/3²+...+1/2004²)
=1-[1/(2.2)+1/(3.3)+...+1/(2004.2004)]
Ta thấy:
1/(2.2)>1/(2.3)
1/(3.3)>1/(3.4)
...
1/(2004.2004)>1/(2004.2005)
Cộng từng vế của các bất đẳng thức trên ta được:
1/(2.2)+1/(3.3)+...+1/(2004.2004) > 1/(2.3)+1/(3.4)+...+1/(2004.2005) = 1/(3.2)+1/(4.3)+...+1/(2005.2004) 
= (3-2)/(3.2)+(4-3)/(4.3)+...+(2005-2004)/(2005.2004)
=3/(3.2)-2/(3.2)+4/(4.3)-3/(4.3)+...+2005/(2005.2004)-2004/(2005.2004)
=1/2-1/3+1/3-1/4+...+1/2004-1/2005
=1/2-1/2005
=2003/4010
=> B>1-2003/4010=2007/4010>2007/4022028=1/2004
Hay B>1/2004
tích nha

cảm ơn bạn nhưng bạn đọc sai đề rồi kìa

a, \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)

Ta có: \(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)

\(\Rightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

Vậy...

b, Đặt A là tên của tổng trên

Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)

Đặt B là biêu thức trong ngoặc

Ta có: \(1=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(\Rightarrow B< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow B< 2-\frac{1}{50}< 2\)

Thay B vào A ta được:

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

c, Đặt C = \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)

\(C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)

\(C=\left(1+1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)

\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)

Đặt D là biểu thức trong ngoặc

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow D< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

\(\Rightarrow-D>-1\)

=>\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1=48\)

Vậy C > 48

Ta gọi tổng trên là B.

Ta luôn có : \(\frac{1}{2^2}>0\)

\(\Rightarrow B>0\left(1\right)\)

Ta có : 

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{44.45}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{44}-\frac{1}{45}\)

\(B< 1-\frac{1}{45}\)

\(B< \frac{44}{45}\)

\(\Rightarrow B< 1\left(2\right)\)

Từ (1) và (2 ) 

\(\Rightarrow0< B< 1\)

=> Tổng B không nguyên

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{44.45}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< \frac{44}{45}< 1\)

\(\Leftrightarrow0< \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)

=> Tổng trên không nguyên

1) Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

    \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

      \(=1-\frac{1}{100}< 1\)

Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 1\)

1) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 1\)

2)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)

 \(\frac{1}{2^2}< \frac{1}{1.2}\)

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

\(...\)

\(\frac{1}{99^2}< \frac{1}{98.99}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)

\(1-\frac{1}{99}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< \frac{99}{100}< 1\)

\(\frac{98}{99}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)

\(\frac{1}{2.3}< \frac{1}{2^2}\)

\(\frac{1}{3.4}< \frac{1}{3^2}\)

\(...\)

\(\frac{1}{99.100}< \frac{1}{99^2}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)

\(\frac{1}{2}-\frac{1}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)

\(\frac{49}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< 1\)

Ta có 

B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)    \(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)= \(1-\frac{1}{8}< 1\)

đúng không đấy mình thấy cứ sai sai thế nào ý

mũ hai nhưng \(8^2\) mới lập lại 1 lần

Thank nhưng mình cần kiểm tra lại 

4S=\(\dfrac{4}{2^2}-\dfrac{4}{2^4}+\dfrac{4}{2^6}-...+\dfrac{4}{2^{4n-2}}-\dfrac{4}{2^{4n}}+...+\dfrac{4}{2^{2002}}-\dfrac{4}{2^{2004}}\)

4S=1-\(\dfrac{1}{2^2}+\dfrac{1}{2^4}-,...-\dfrac{1}{2^{2002}}\)

4S+S=1-\(\dfrac{1}{2^{2004}}\)

5S=\(\dfrac{2^{2004}-1}{2^{2004}}\)<1

\(\Rightarrow\)5S<1 hay S<\(\dfrac{1}{5}\)=0,2(đpcm)