Question-63904

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

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

\frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} = \frac{3^{k}}{3^{k} - 2^{k}} - \frac{3^{k + 1}}{3^{k + 1} - 2^{k + 1}}

\underset{k = 1}{\sum^{n}} = 3 - \frac{3^{n + 1}}{3^{n + 1} - 2^{n + 1}} = 3 - \frac{1}{1 - {(\frac{2}{3})}^{n + 1}}

as n \to \infty \Rightarrow {(\frac{2}{3})}^{n + 1} \to 0

∴ \underset{k = 1}{\sum^{\infty}} \frac{6^{k}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1})} \to 3 - 1 = 2

please check