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Question Number 211129 by depressiveshrek last updated on 29/Aug/24

Commented by depressiveshrek last updated on 29/Aug/24

Find the sum of this series

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{this}\:\mathrm{series} \\ $$

Answered by mr W last updated on 29/Aug/24

=Σ_(n=1) ^∞ ((9(n+1)×3^n ×((√u))^n )/((x^n +y^n )^2 ((z)^(1/3) )^n )) =9Σ_(n=1) ^∞ (((n+1))/((x^n +y^n )^2 ))(((3(√u))/( (z)^(1/3) )))^n =9Σ_(n=1) ^∞ (((n+1))/((a^n +b^n )^2 )) with a=x(√((z)^(1/3) /(3(√u)))), b=y(√((z)^(1/3) /(3(√u)))) .......

$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{9}\left({n}+\mathrm{1}\right)×\mathrm{3}^{{n}} ×\left(\sqrt{{u}}\right)^{{n}} }{\left({x}^{{n}} +{y}^{{n}} \right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{{z}}\right)^{{n}} } \\ $$$$=\mathrm{9}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{1}\right)}{\left({x}^{{n}} +{y}^{{n}} \right)^{\mathrm{2}} }\left(\frac{\mathrm{3}\sqrt{{u}}}{\:\sqrt[{\mathrm{3}}]{{z}}}\right)^{{n}} \\ $$$$=\mathrm{9}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{1}\right)}{\left({a}^{{n}} +{b}^{{n}} \right)^{\mathrm{2}} }\:{with}\:{a}={x}\sqrt{\frac{\sqrt[{\mathrm{3}}]{{z}}}{\mathrm{3}\sqrt{{u}}}},\:{b}={y}\sqrt{\frac{\sqrt[{\mathrm{3}}]{{z}}}{\mathrm{3}\sqrt{{u}}}} \\ $$$$....... \\ $$

Commented by mr W last updated on 29/Aug/24

unsolvable or solvable?

$${unsolvable}\:{or}\:{solvable}? \\ $$

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