Question-209800

Question Number 209800 by depressiveshrek last updated on 22/Jul/24

Commented by depressiveshrek last updated on 22/Jul/24

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

Answered by mr W last updated on 22/Jul/24

=Σ_(n=1) ^∞ (1/3^2 )(((3(√2))/(xy^2 (z)^(1/3) )))^n (n−1) =(1/9)Σ_(n=1) ^∞ (n−1)k^n with k=((3(√2))/(xy^2 (z)^(1/3) )) assume ∣k∣<1, otherwise it′s not convergent. Σ_(n=1) ^∞ x^(n−1) =(1/(1−x)) Σ_(n=1) ^∞ (n−1)x^(n−2) =(1/((1−x)^2 )) Σ_(n=1) ^∞ (n−1)x^n =(x^2 /((1−x)^2 ))=((1/(1−(1/x))))^2 set x=k: Σ_(n=1) ^∞ (n−1)k^n =((1/(1−(1/k))))^2 =((1/(1−((xy^2 (z)^(1/3) )/(3(√2))))))^2 (1/9)Σ_(n=1) ^∞ (n−1)k^n =((1/(3−((xy^2 (z)^(1/3) )/( (√2))))))^2 =(2/((3(√2)−xy^2 (z)^(1/3) )^2 )) ✓

$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\left(\frac{\mathrm{3}\sqrt{\mathrm{2}}}{{xy}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{{z}}}\right)^{{n}} \left({n}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}−\mathrm{1}\right){k}^{{n}} \:\:\:{with}\:{k}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{{xy}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{{z}}} \\ $$$${assume}\:\mid{k}\mid<\mathrm{1},\:{otherwise}\:{it}'{s}\:{not} \\ $$$${convergent}. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}−\mathrm{1}\right){x}^{{n}} =\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\right)^{\mathrm{2}} \\ $$$${set}\:{x}={k}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}−\mathrm{1}\right){k}^{{n}} =\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{k}}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{{xy}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{{z}}}{\mathrm{3}\sqrt{\mathrm{2}}}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}−\mathrm{1}\right){k}^{{n}} =\left(\frac{\mathrm{1}}{\mathrm{3}−\frac{{xy}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{{z}}}{\:\sqrt{\mathrm{2}}}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}}{\left(\mathrm{3}\sqrt{\mathrm{2}}−{xy}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{{z}}\right)^{\mathrm{2}} }\:\checkmark \\ $$

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