1-1-pi-2-n-2-1-n-1-n-1-2-1-pi-2-n-2-1-e-n-1-n-1-A-1-lt-2-B-1-2-C-1-gt-2-

Question Number 156900 by MathSh last updated on 16/Oct/21

Ω_{1} = 1 - \frac{π}{2} + \underset{n = 2}{\sum^{\infty}} {(- \frac{1}{π})}^{n} \cdot \frac{1}{n + 1}

Ω_{2} = 1 - \frac{π}{2} + \underset{n = 2}{\sum^{\infty}} {(- \frac{1}{e})}^{n} \cdot \frac{1}{n + 1}

A) Ω_{1} < Ω_{2} B) Ω_{1} = Ω_{2} C) Ω_{1} > Ω_{2}