Đặt A =\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2008}{3^{2008}}\)

Suy ra 3A = \(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2008}{3^{2007}}\)=> 2A = 3A - A = \(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2008}{3^{2007}}-\frac{1}{3}-\frac{2}{3^2}-\frac{3}{3^3}-...-\frac{2008}{3^{3008}}\)= \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}-\frac{2008}{3^{2008}}\)

= \(\frac{3}{2}-\frac{1}{2.3^{2007}}\)Suy ra A = \(\frac{3}{4}-\frac{1}{8.3^{2007}}\)<\(\frac{3}{4}\)(ĐPCM)

có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 +  2^10]

Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]

Q = 2 . 3+2^3 .3 +... + 2^9 .3

Q = 3. [ 2 + 2^3 +... + 2^9]

Vậy Q chia hết cho 3

Ta có:

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

\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

\(< \left(1+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Áp dụng vào bài toán ta được

\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2009\sqrt{2008}}\)

\(=2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)\)

\(=2\left(1-\frac{1}{\sqrt{2009}}\right)< 2\)

To quá

Gọi a là tử số, b là mẫu số của phân số A

a = \(\frac{2008}{1}\)+ \(\frac{2007}{2}\)+ \(\frac{2006}{3}\)+ ... + \(\frac{1}{2008}\)

Dãy số a có (2008 - 1)  : 1 + 1 = 2008 số. Và a = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2) 

b = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{2009}\)

Dãy số b có (2009 - 2) : 1 + 1 = 2008 số. Và b = (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)

A = [ ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)] : [ (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)] = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) :  (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) 

A = \(\frac{\text{2008 x2008 + 1}}{2008}\)x \(\frac{2x2009+2}{2x2009}\)

A = 2008

Ap dung \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Ta co \(P< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2008}}\right)\)  

=> \(P< 2\left(1-\frac{1}{\sqrt{2008}}\right)< 2.1=2\)

Suy ra P khong phai so nguyen to

\(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...+\left(1+\frac{1}{2007}\right)+\left(1+\frac{1}{2008}\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)