Question Number 30049 by abdo imad last updated on 15/Feb/18
$${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{3}^{{n}} }\:. \\ $$
Commented by prof Abdo imad last updated on 16/Feb/18
![we have proved that Σ_(n=0) ^∞ (n+1)x^n = (1/((1−x)^2 )) for ∣x∣<1 due to uniform convergence on [0,1] by derivation Σ_(n=1) ^∞ n(n+1)x^(n−1) =(d/dx)((1−x)^(−2) ) = (−2)(−1)(1−x)^(−3) = (2/((1−x)^3 )) ⇒ Σ_(n=1) ^∞ n(n+1)x^n = ((2x)/((1−x)^3 )) and for x=(1/3) we get Σ_(n=1) ^∞ ((n(n+1))/3^n ) = ((2/3)/(((2/3))^3 )) = (1/(((2/3))^2 )) = (9/4) =.](https://www.tinkutara.com/question/Q30134.png)
$${we}\:{have}\:{proved}\:{that}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}+\mathrm{1}\right){x}^{{n}} \:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${for}\:\mid{x}\mid<\mathrm{1}\:{due}\:{to}\:{uniform}\:{convergence}\:{on}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$${by}\:{derivation}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\left({n}+\mathrm{1}\right){x}^{{n}−\mathrm{1}} =\frac{{d}}{{dx}}\left(\left(\mathrm{1}−{x}\right)^{−\mathrm{2}} \right) \\ $$$$=\:\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)\left(\mathrm{1}−{x}\right)^{−\mathrm{3}} =\:\frac{\mathrm{2}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} {n}\left({n}+\mathrm{1}\right){x}^{{n}} =\:\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:{and}\:{for}\:{x}=\frac{\mathrm{1}}{\mathrm{3}}\:{we}\:{get} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{3}^{{n}} }\:=\:\frac{\frac{\mathrm{2}}{\mathrm{3}}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{9}}{\mathrm{4}}\:=. \\ $$$$ \\ $$