Question-216703

Question Number 216703 by LMKhit last updated on 16/Feb/25

Answered by issac last updated on 16/Feb/25

\sqrt{1 - x^{2}} = \frac{\sqrt{π}}{2} \underset{h = 0}{\sum^{\infty}} \frac{{(- 1)}^{h}}{h! (\frac{1}{2} - h)!} x^{2 h}, ∣ x ∣\leq 1

because r = \underset{n \to \infty}{\lim i n f} ∣ \frac{a_{n + 1}}{a_{n}} ∣\leq 1

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