a_2=\frac{1}{ln2^{(2-1)}*ln2^{(2)}}=\frac{1}{2ln^22}=\frac{1}{ln^22}\frac{1}{(1*2)}, a_3=\frac{1}{ln2^{(2-1)}*ln2^{(2)}}+\frac{1}{ln2^{(3-1)}*ln2^{3} }=\frac{1}{2ln^22}+\frac{1}{6ln^22}=\frac{1}{ln^22}[\frac{1}{(1*2)}+\frac{1}{(2*3)}]