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Question Number 140966 by mnjuly1970 last updated on 14/May/21

$$.......nice......calculus.....$$$$if\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\sqrt{cos\left(n\pi \right)}}{\left(2n\right)!!}=\omega$$$$thenRe\left(\omega \right):=??$$

Answered by Dwaipayan Shikari last updated on 14/May/21

$$cos\left(n\pi \right)={(-1)}^n$$$$\sqrt{cos\left(n\pi \right)}=i^n$$$$\frac{1}{\left(2n\right)!!}=\frac{1}{2.4.6.8.10.12...2n}=\frac{1}{2^{n}n!}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{i^n}{2^{n}n!}=e^{i/2}=cos\left(1/2\right)+isin\left(1/2\right)$$$$Re\left(\omega \right)=cos\left(\frac{1}{2}\right)$$

Commented by mnjuly1970 last updated on 14/May/21

$$thanksalotmrpayan...$$

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