Let \(A=\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+...+\dfrac{1}{101}\)

We have \(A< \dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}< 1\)

Cause A always \(>0\), but \(A< 1\Rightarrow A\notin N\)

Why is A smaller than \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)?

I can't understand.A has : (101 - 5) : 2 + 1 = 49 (terms) but the sum you gave has only 9 terms.The solution isn't logical