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Question Number 63904 by naka3546 last updated on 11/Jul/19

Commented by Prithwish sen last updated on 13/Jul/19

$$\frac{6^{\mathrm{k}}}{(3^{\mathrm{k}}-2^{\mathrm{k}})(3^{\mathrm{k}+1}-2^{\mathrm{k}+1})}=\frac{3^{\mathrm{k}}}{3^{\mathrm{k}}-2^{\mathrm{k}}}-\frac{3^{\mathrm{k}+1}}{3^{\mathrm{k}+1}-2^{\mathrm{k}+1}}$$$$\underset{\mathrm{k}=1}{\sum ^{\mathbf{n}}}=3-\frac{3^{\mathrm{n}+1}}{3^{\mathrm{n}+1}-2^{\mathrm{n}+1}}=3-\frac{1}{1-{\left(\frac{2}{3}\right)}^{\mathrm{n}+1}}$$$$\mathbf{as}\mathbf{n}\to \mathrm{\infty }\Rightarrow {\left(\frac{2}{3}\right)}^{\mathbf{n}+1}\to 0$$$$\therefore \underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{6^{\mathrm{k}}}{(3^{\mathrm{k}}-2^{\mathrm{k}})(3^{\mathrm{k}+1}-2^{\mathrm{k}+1})}\to 3-1=2$$$$\mathrm{please}\mathrm{check}$$

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