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Ta có:
$S=\dfrac{5}{1\cdot2\cdot3}+\dfrac{8}{2\cdot3\cdot4}+\dfrac{11}{3\cdot4\cdot5}+\cdots+\dfrac{6026}{2008\cdot2009\cdot2010}$
Nhận thấy tử số có dạng $3n+2$, nên:
$S=\sum_{n=1}^{2008}\dfrac{3n+2}{n(n+1)(n+2)}$
Ta có: $\dfrac{3n+2}{n(n+1)(n+2)}=\dfrac{1}{n(n+1)}+\dfrac{2}{(n+1)(n+2)}$
Do đó:
$S=\left(\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots+\dfrac1{2008\cdot2009}\right)$
$+2\left(\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{2009\cdot2010}\right)$
Mà: $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$
Nên: $S=\left(1-\dfrac1{2009}\right)+2\left(\dfrac12-\dfrac1{2010}\right)$
$=1-\dfrac1{2009}+1-\dfrac1{1005}$
$=2-\dfrac1{2009}-\dfrac1{1005}$
Vì: $\dfrac1{2009}+\dfrac1{1005}>0$
Nên $S<2$
Tổng quát: \(\frac{2}{\left(a-1\right)a\left(a+1\right)}=\frac{1}{\left(a-1\right).a}-\frac{1}{a\left(a+1\right)}\)
Ta có: \(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+.....+\frac{2}{2013.2014.2015}\)
\(S=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+.....+\left(\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)
\(S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\)
\(S=\frac{1}{1.2}-\frac{1}{2014.2015}=\frac{1}{2}-\frac{1}{2014.2015}<\frac{1}{2}\)
Vậy....................
S=(2/1.2-2/2.3)+(2/2.3-2/3.4)+(2/3.4-2/4.5)+...........+(2/2013.2014-2/2014-2/2015)
S=(2/1.2-2/2014.2015):2
S=1-2/2014.2/2015
--> S>1/2