Question Number 66680 by mathmax by abdo last updated on 18/Aug/19
$${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}^{{n}} }{\mathrm{3}^{{n}} \left(\mathrm{2}{n}^{\mathrm{3}} \:+{n}^{\mathrm{2}} −\mathrm{5}{n}\:+\mathrm{2}\right)} \\ $$
Commented by Mohamed Amine Bouguezzoul last updated on 18/Aug/19
$${it}\:{cannot}\:{be}\:{with}\:{n}=\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
$${sorry}\:{the}\:{Q}\:{is}\:{calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{2}^{{n}} }{\mathrm{3}^{{n}} \left(\mathrm{2}{n}^{\mathrm{3}} \:+{n}^{\mathrm{2}} −\mathrm{5}{n}\:+\mathrm{2}\right)} \\ $$