Câu 1: 

\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\)

\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\)

\(=\dfrac{5}{4}\cdot\dfrac{4n+3-3}{3\left(4n+3\right)}=\dfrac{5}{4}\cdot\dfrac{4n}{3\left(4n+3\right)}=\dfrac{5n}{3\left(4n+3\right)}\)

Câu 2: 

\(=\dfrac{3}{5}\left(\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\right)\)

\(=\dfrac{3}{5}\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right)\)

\(=\dfrac{3}{5}\cdot\dfrac{5n+4-9}{9\left(5n+4\right)}=\dfrac{3}{5}\cdot\dfrac{5\left(n-1\right)}{9\left(5n+4\right)}=\dfrac{n-1}{3\left(5n+4\right)}< \dfrac{1}{15}\)

a/ \(\frac{1}{n\left(n-1\right)\left(n+1\right)}=\frac{1}{n^3-n}>\frac{1}{n^3}\)

b/ \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n^3+3n^2+2n}< \frac{1}{n^3}\)

c/ Ap dụng câu b ta được

\(\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2006^3}>\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2006.2007.2008}\)

\(=\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2006.2007}-\frac{1}{2007.2008}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{2007.2008}\right)>\frac{1}{12}>\frac{1}{15}\)

chịu

Ta có : 1/(n-1).n.(n+1) = 1/n.[(n-1).(n+1)] = 1/n.(n^2-1)

Vì n là số nguyên > 1 => n > 0 ; n^2-1 > 0 và n^2-1 < n^2

=> 1/n^2 < 1/n^2-1

=> 1/n.n^2 < 1/n.(n^2+1)

=> 1/n^3 < 1/n.(n-1).(n+1)

=> ĐPCM

Tk mk nha

help mình vs plz

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