\(100-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)

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

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

\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)

\(\Rightarrow\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}< S< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)

\(\frac{3}{14}\times5< S< \frac{3}{10}\times5\Rightarrow\frac{15}{14}< S< \frac{3}{2}\)

mà \(\frac{15}{14}>1;\frac{3}{2}< 2\Rightarrow1< S< 2\)

1/100² + 1/101² + ... + 1/199² < 1/99.100 + 1/100.101 + ... + 1/198.199 = 1/99 - 1/100 + 1/100 - 1/101 + ... + 1/198 - 1/199 = 1/99 - 1/199

\(\Rightarrow\)Vậy 1/100² + 1/101² + ... + 1/199² < 1/99 (vì 1/99 đã lớn hơn 1/99 - 1/199 rồi mà G lại còn bé hơn 1/99 - 1/199 nữa)

1/100² + 1/101² + ... + 1/199² > 1/100.101 + ... + 1/199.200 = 1/100 - 1/101 + ... + 1/199 - 1/200 = 1/100 - 1/200 = 1/200

\(\Rightarrow\)Vậy 1/100² + 1/101² + ... + 1/199²  > 1/200

?

Gỉa sử\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

=>\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=>\(A< 1-\dfrac{1}{100}\)
=>\(A< \dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\)
=>A<1
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

1/2^2 + 1/3^2 + ...+ 1/100^2 
Ta có : 1/2^2 < 1/1.2
1/3^2 < 1/2.3
...
1/100^2 < 1/99.100
=> 1/2^2 + ...+1/100^2 < 1/1.2+1/2.3+...+1/99.100
                                        = 1 - 1/2+1/2-1/3+1/3+...+1/99-1/100
                                        =  1 - 1/100 <1
-> 1/2^2 + ...+1/100^2 < 1

Ta có:

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

M=\(\frac{1.3....99}{2.4....100}\)

Lại có:

N=\(\frac{2}{3}.\frac{4}{5}....\frac{100}{101}\)

N=\(\frac{2.4....100}{3.5....101}\)

\(\Rightarrow\)M.N=\(\frac{1.2.3......99.100}{2.3.4......100.101}\)

\(\Rightarrow\)M.N=\(\frac{1}{101}\)

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

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

....

\(\frac{1}{100^2}=\frac{1}{100.100}<\frac{1}{99.100}\)

do đó \(A<\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}<1\)

=>A<1

sẽ là 1/4+1/9+1/16........tổng sẽ ko lớn hơn 1

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