Ta thấy:

1/2*2<1/1*2)vì 2*2>1*2).

1/3*3<1/2*3(vì 3*3>2*3).

...

1/8*8<1/7*8(vì 8*8>7*8).

=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.

=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8.

=>B<1-1/8.

=>B<7/8.

Mà 7/8<1.

=>B<1.

Vậy B<1(đpcm).

\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(\Rightarrow1-\frac{1}{8}< 1\)

=>B<1

\(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6.}+\frac{1}{7.7}+\frac{1}{8.8}\)\(=\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(B=1-\frac{1}{8}\)

\(\Rightarrow B< 1\left(ĐPCM\right)\)

1 + 2 x 2 + 3 x 3 + 4 x 4 + 5 x 5 + 6 x 6 + 7 x 7 + 8 x 8 + 9 x 9 + 10 x 10

= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100

= 385

1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10

= 1+4+9+16+25+36+49+64+81+100

=(81+9)+(64+16)+(49+1)+)36+4)+25+100

=90+80+50+40+25 +100

=385

Đặt T = 1 + 2 x 2 + ... + 10 x 2 = 385

ta có: \(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2007}{2^{2007}}\)

\(\Rightarrow\frac{1}{2}S=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+...+\frac{2007}{2^{2008}}\)

\(\Rightarrow S-\frac{1}{2}S=\frac{1}{2}+\left(\frac{2}{2^2}-\frac{1}{2^2}\right)+\left(\frac{3}{2^3}-\frac{2}{2^3}\right)+\left(\frac{4}{2^4}-\frac{3}{2^4}\right)+...+\left(\frac{2007}{2^{2007}}-\frac{2006}{2^{2007}}\right)-\frac{2007}{2^{2008}}\)

\(\frac{1}{2}S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\)

Gọi \(Q=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2007}}\)

\(\Rightarrow\frac{1}{2}Q=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2008}}\)

\(\Rightarrow Q-\frac{1}{2}Q=\frac{1}{2}-\frac{1}{2^{2008}}\)

\(\Rightarrow\frac{1}{2}Q=\frac{1}{2}-\frac{1}{2^{2008}}\)

\(Q=\left(\frac{1}{2}-\frac{1}{2^{2008}}\right):\frac{1}{2}=1-\frac{1}{2^{2007}}\)

Thay Q vào S, ta có:

\(\frac{1}{2}S=1-\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\)

\(\Rightarrow S=\left(1-\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\right):\frac{1}{2}\)

\(S=2-\frac{1}{2^{2006}}-\frac{2007}{2^{2007}}< 2\)

\(\Rightarrow S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2007}{2^{2007}}< 2\)

Ta có:

1/5×5 < 1/4×5

1/6×6 < 1/5×6

1/7×7 < 1/6×7

.........

1/100×100 < 1/99×100

=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 < 1/4×5 + 1/5×6 + 1/6×7 +.....+ 1/99×100

                                      = 1/4-1/5 + 1/5-1/6 + 1/6-1/7 +......+ 1/99-1/100

                                    = 1/4-1/100 < 1/4  

=> 1/5×5 + 1/6×6+1/7×7 +...+1/100×100<1/4  (1)

Lại có:

1/5×5 > 1/6×7

1/6×6 > 1/7×8

1/7×7 > 1/8×9

........

1/100×100 > 1/101×102

=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 > 1/5×6 + 1/6×7 + 1/7×8  +.....+1/100×101

                                   = 1/5-1/6 + 1/6-1/7 + 1/7-1/8 +.....+ 1/100 - 1/101

                                   = 1/5 - 1/101 > 1/5 - 1/30 = 1/6

=> 1/5×5 + 1/6×6 +1/7×7 +.....+ 1/100×100>1/6 (2)

Từ (1) và (2)

=> 1/6 < 1/5×5 +1/6×6+ 1/7×7 +...+1/100×100<1/4

Đặt \(A=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)

Có \(\frac{1}{5.5}< \frac{1}{4.5};\frac{1}{6.6}< \frac{1}{5.6};...;\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)(1)

Lại có :\(\frac{1}{5.5}>\frac{1}{5.6};\frac{1}{6.6}>\frac{1}{6.7};...;\frac{1}{100.100}>\frac{1}{100.101}\)

\(\Rightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\left(2\right)\)

Từ (1) và (2) \(\RightarrowĐCCM\)

Đặt \(X=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)

\(\Rightarrow X< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(\Rightarrow X=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow X< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

\(\Rightarrow X< \frac{1}{4}\left(1\right)\)

\(\Rightarrow X>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)

\(\Rightarrow X>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow X>\frac{1}{5}-\frac{1}{101}>\frac{1}{5}-\frac{1}{30}=\frac{1}{6}\)

\(\Rightarrow X>\frac{1}{6}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{6}< X< \frac{1}{4}\)

5/8+3/5x5/12=

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

Với n > 2 thì \(\frac{n}{2^n}=\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}\)

\(\frac{n+1}{2^{n-1}}=\frac{n+1}{2^n:2}=\frac{n+1}{\frac{2^n}{2}}=\frac{2^{\left(n+1\right)}}{2^n}\)

\(\frac{n+1}{2^{n-1}}-\frac{n+2}{2^n}=\frac{2^{n+2}}{2^n}-\frac{n+2}{2^n}\)

\(=\frac{2^{n+2}-n-2}{2^n}\)

\(=\frac{n}{2^n}\)

\(\Leftrightarrow S=\frac{1}{2}+\left(\frac{2+1}{2^{2-1}}-\frac{2+2}{2^2}\right)+.....+\frac{2016+1}{2^{2015}}-\frac{2018}{2^{2016}}\)

\(=\frac{2017+1}{2^{2016}}-\frac{2019}{2^{2017}}\)

\(S=\frac{1}{2}+\frac{3}{2}-\frac{2019}{2017}\)

\(S=2-\frac{2019}{2017}\)

\(\Leftrightarrow S=2-\frac{2019}{2017}< 2\)

Hay \(S< 2\)

\(M=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)

\(\Rightarrow2M=2+1+\frac{1}{2}+...+\frac{1}{2^{2015}}\)

\(\Rightarrow2M-M=2-\frac{1}{2^{2016}}< 2\)

=>ĐPCM

ta có

M = 1+1/2 +...+1/2²⁰¹⁶

2M=2+1+1/2+...+1/2²⁰¹⁴+1/2²⁰¹⁵

2M-M=(2+1+1/2+...+1/2²⁰¹⁵)-(1+1/2 +...+1/2²⁰¹⁶)

M=2-1/2²⁰¹⁶<2

Vậy M<2