a>

\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000

ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )

1/100^2<1/2

=>A<1

oke

CMR A = 1/5 + 1/6 + 1/7 + ... + 1/17 < 2

A = 1/5 + 1/6 + 1/7 + ... + 1/17

Vì 1/6 < 1/7 < 1/8 < 1/9 < 1/5 và 1/10 < 1/11 < 1/12 < 1/13 < 1/14 <1/15 < 1/16 < 1/17 < 1/8 nên:

A = (1/5 + 1/6 + 1/7 + 1/8 + 1/9) + (1/10 + 1/11 + 1/12 + 1/13 + 1/14+ 1/15 + 1/16 + 1/17)

A < (1/5 + 1/5 + 1/5 + 1/5 + 1/5) + (1/8 + 1/8 + 1/8 + 1/8 + 1/8+1/8 + 1/8 + 1/8)

A < 1 + 1

A < 2

Vậy: A < 2 (đpcm)

A= \(\frac{1}{2}\) + \(\frac{1}{2^2}\) + \(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)

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

\(\Rightarrow\) 2A - A = ( \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\) ) -

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

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

Vậy: A < 1
\(\frac{1}{2}\)

B= \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

= 2. \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

= 2. \(\left(\frac{1}{1}-\frac{1}{100}\right)\) = \(\frac{99}{50}\)

\(\Rightarrow\) B = \(\frac{99}{50}\) < \(\frac{100}{50}\) = 2

Vậy: B < 2

Ta có :

S= 1/51 +1/52 +..+1/100

Vì 1/51>1/52>...>1/100 

=> S >1/100 * 50 =1/2 (1)

Vì 1/100 <1/99<...<1/51<1/50

=> S < 1/50 * 50=1 (2)

Từ (1),(2) => 1/2 < S<1

P=1/2^2+1/2^3+...+1/2^2018 

2P=1/2 +1/2^2 +...+1/2^2017

=> 2P-P= (1/2 +1/2^2 +...+1/2^2017)-(1/2^2+1/2^3+...+1/2^2018 )

=> P=1/2 -1/2^2018 <1/2 <3/4

Ta có: \(\frac{1}{51}>\frac{1}{100};\frac{1}{52}>\frac{1}{100};...;\frac{1}{100}=\frac{1}{100}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{1}{100}.50=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}\)

Ta có \(\frac{1}{51}< \frac{1}{50};\frac{1}{52}< \frac{1}{50};...;\frac{1}{100}< \frac{1}{50}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{50}.50=1\)

\(\Rightarrow S< 1\)

`Answer:`

 \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)

a) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)

\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)

\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)

b) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)

\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)

\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)

\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)

\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)

Ta có : \(\frac{1}{2^2}>\frac{1}{2.3}\)

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

              ..................

           \(\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{100.101}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{100^2}>\frac{1}{2}-\frac{1}{101}>\frac{1}{6}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{4}\)

Nên : \(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{100^2}>\frac{1}{4}\) (nên đề vô lý)