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S-n-12-4-2-3-2-4-2-3-2-12-2-4-2-3-2-4-3-3-3-12-3-4-2-3-2-4-4-3-4-12-n-4-2-3-2-4-n-1-3-n-1-lim-n-S-n-




Question Number 160604 by cortano last updated on 03/Dec/21
  S_n = ((12)/((4^2 −3^2 )(4^2 −3^2 )))+((12^2 )/((4^2 −3^2 )(4^3 −3^3 )))+((12^3 )/((4^2 −3^2 )(4^4 −3^4 )))+...+((12^n )/((4^2 −3^2 )(4^(n+1) −3^(n+1) )))   lim_(n→∞)  S_n  = ?
$$\:\:\mathrm{S}_{\mathrm{n}} =\:\frac{\mathrm{12}}{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)}+\frac{\mathrm{12}^{\mathrm{2}} }{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}} \right)}+\frac{\mathrm{12}^{\mathrm{3}} }{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \right)}+…+\frac{\mathrm{12}^{\mathrm{n}} }{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{n}+\mathrm{1}} −\mathrm{3}^{\mathrm{n}+\mathrm{1}} \right)} \\ $$$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{S}_{\mathrm{n}} \:=\:? \\ $$

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