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Question Number 123763 by mnjuly1970 last updated on 28/Nov/20

$$...nicecalculus...$$$$\mathcal{E}valuate::\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{2n+1}=???$$

Answered by Snail last updated on 28/Nov/20

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^{2n+1}}{2n+1}={tan}^{-1}\left(x\right)$$$$$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{2n+1}={tan}^{-1}\left(1\right)=\pi /4$$

Commented by mnjuly1970 last updated on 28/Nov/20

$$thankyousomuch$$

Answered by Dwaipayan Shikari last updated on 28/Nov/20

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{2n+1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$$$$Anotherway$$$$cotx=\frac{1}{x}-\frac{1}{\pi -x}+\frac{1}{\pi +x}-\frac{1}{2\pi -x}+\frac{1}{2\pi +x}+...$$$$x=\frac{\pi }{4}\Rightarrow 1=\frac{4}{\pi }-\frac{4}{3\pi }+\frac{4}{5\pi }-\frac{4}{7\pi }+...$$$$\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$$

Commented by mnjuly1970 last updated on 28/Nov/20

$$verynicebravosirdwaipayan..$$

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