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Question Number 205774 by mnjuly1970 last updated on 30/Mar/24

$$$$$$-------$$$$\mathrm{\Omega }=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(-1)}^n-n}=?$$$$-------$$

Answered by Berbere last updated on 30/Mar/24

$$\mathrm{\Omega }=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{1-2n}-\frac{1}{-2n-2})$$$$=\sum _{n⩾0}(\frac{1}{1-2n}+\frac{1}{2(n+1)})=\sum _{n⩾0}(\frac{1}{2(n+1)}-\frac{1}{2n-1})$$$$=\frac{3}{2}-\sum _{n⩾1}\frac{1}{2n+1+1}-\frac{1}{2n-2+1}$$$$=\frac{3}{2}+\sum _{n⩾1}({\int }_0^1{x}^{2n+1}-x^{2n-2}dx)$$$$S\left(N\right)=\sum _{n⩾0}{\int }_0^1\frac{x^3}{1-x^2}-\frac{1}{1-x^2}=\frac{3}{2}+{\int }_0^1\frac{x^3-1}{(1-x)(1+x)}dx$$$$=\frac{3}{2}-{\int }_0^1\frac{x^2+x+1}{x+1}=\frac{3}{2}-{\int }_0^1\frac{x(x+1)+1}{x+1}dx$$$$=\frac{3}{2}-{\int }_0^{1}x+\frac{1}{1+x}dx=1-ln\left(2\right)$$$$\mathrm{\Omega }=1-ln\left(2\right)$$$$$$

Commented by mnjuly1970 last updated on 30/Mar/24

$$sogoodsolution$$

Answered by MM42 last updated on 30/Mar/24

$$\mathrm{\Omega }=\left(\frac{1}{1}\right)+\left(\frac{-1}{-1-1}\right)+\left(\frac{1}{1-2}\right)+\left(\frac{-1}{-1-3}\right)+\left(\frac{1}{1-4}\right)+\left(\frac{-1}{-1-5}\right)+\left(\frac{1}{1-6}\right)+(\frac{-1}{-1-7)}...$$$$1+\frac{1}{2}-1+\frac{1}{4}-\frac{1}{3}+\frac{1}{6}-\frac{1}{5}+\frac{1}{8}-...$$$$=1-(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+..)$$$$=1-ln2✓$$$$$$

Commented by mnjuly1970 last updated on 30/Mar/24

$$thanksalot$$

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