Question

Sum to infinity the following series :
i) $1+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+...$
ii) $1-\frac{3}{2}+\frac{5}{4}-\frac{7}{8}+....$
iii) $1+4x^2+7x^4+10x^6+....,|x|$

Answer 1

$S=1+\frac{3}{2}+\frac{5}{2^2}+.............$

$S\frac{1}{2}=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+.............$

$S-S\frac{1}{2}=1+\frac{3}{2}+\frac{5}{2^2}+.............-\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+.............$

$\frac{S}{2}=1+\frac{2}{2}+\frac{2}{2^2}+.............$

$\frac{S}{2}=1+\frac{1}{1-\frac{1}{2}}$

$S=6$

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$S=1-\frac{3}{2}+\frac{5}{4}-\frac{7}{8}+...$

$\frac{S}{2}=\frac{1}{2}-\frac{3}{4}+\frac{5}{8}+...$

$S+\frac{S}{2}=\frac{3}{2}S=1-\frac{2}{2}+\frac{2}{4}-\frac{2}{8}+...$

Double that:

$3S=2-\frac{4}{2}+\frac{4}{4}-\frac{4}{8}+...$

Add those last two together:

$\frac{9}{2}S=2+1-\frac{4}{2}$, and the rest cancels itself out

$\frac{9}{2}S=2+1-2$

$\frac{9}{2}S=1$

$S=\frac{2}{9}$

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$S=1+4x^2+7x^4+10x^6+............$

$S{x}^2=x^2+4x^4+7x^6+10x^8+..........$

$S-S{x}^2=1+3x^2+3x^4+3x^6+............$

$S(1-x^2)=1+3(x^2+x^4+x^6+...........)$

$S(1-x^2)=1+3\left(\frac{x^2}{1-x^2}\right)$

$S=\frac{1}{1-x^2}+3\frac{x^2}{(1-x^2{)}^2}$

$S=\frac{1+2x^2}{(1-x^2{)}^2}$