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Question Number 106860 by Dwaipayan Shikari last updated on 07/Aug/20

$$1-1+\frac{5}{9}-\frac{7}{27}+\frac{9}{81}-\frac{11}{243}+.....$$

Commented by Dwaipayan Shikari last updated on 07/Aug/20

$$1-\frac{3}{3}+\frac{5}{9}-\frac{7}{27}+\frac{9}{81}-\frac{11}{243}+..$$$$S=1-3x+5x^2-7x^3+9x^4-11x^5+..(x=\frac{1}{3})$$$$xS=x-3x^2+5x^3-...$$$$S(1+x)=1-2x+2x^2-2x^3+....$$$$S(1+x)=1-2(x-x^2+x^3-....)$$$$S(1+x)=1-\frac{2x}{1+x}$$$$S=\frac{1-x}{{(1+x)}^2}=\frac{\frac{2}{3}}{\frac{16}{9}}=\frac{3}{8}$$

Commented by bemath last updated on 07/Aug/20

$$\mathrm{cooll}$$

Commented by abdomathmax last updated on 07/Aug/20

$$\mathrm{S}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}(2\mathrm{n}+1)}{3^{\mathrm{n}}}\Rightarrow \mathrm{S}=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{\mathrm{n}{(-1)}^{\mathrm{n}}}{3^{\mathrm{n}}}$$$$+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{3^{\mathrm{n}}}\mathrm{we}\mathrm{have}$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-\frac{1}{3})}^{\mathrm{n}}=\frac{1}{1+\frac{1}{3}}=\frac{3}{4}$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\mathrm{n}{(-\frac{1}{3})}^{\mathrm{n}}=\mathrm{f}(-\frac{1}{3})\mathrm{with}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}$$$$\mathrm{for}\mid \mathrm{x}<1\mathrm{\Sigma }{\mathrm{x}}^{\mathrm{n}}=\frac{1}{1-\mathrm{x}}\Rightarrow \mathrm{\Sigma }{\mathrm{nx}}^{\mathrm{n}-1}=\frac{1}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\mathrm{\Sigma }{\mathrm{nx}}^{\mathrm{n}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}=\mathrm{f}\left(\mathrm{x}\right)\Rightarrow \mathrm{f}(-\frac{1}{3})=\frac{-1}{3{(1+\frac{1}{3})}^2}$$$$=\frac{-1}{3{\left(\frac{4}{3}\right)}^2}=-\frac{3}{16}\Rightarrow \mathrm{S}=2(-\frac{3}{16})+\frac{3}{4}$$$$=-\frac{3}{8}+\frac{6}{8}=\frac{3}{8}$$$$$$

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