n-1-6-n-3-n-2-n-3-n-1-2-n-1-

Question Number 125256 by 676597498 last updated on 09/Dec/20

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

Answered by mathmax by abdo last updated on 09/Dec/20

let u_{n} = \frac{6^{n}}{(3^{n} - 2^{n}) (3^{n + 1} - 2^{n + 1})} \Rightarrow u_{n} = \frac{3^{n} . 2^{n}}{3^{n} (1 - {(\frac{2}{3})}^{n}) 3^{n + 1} (1 - {(\frac{2}{3})}^{n + 1})}

= \frac{2^{n}}{3^{n + 1} (1 - {(\frac{2}{3})}^{n}) (1 - {(\frac{2}{3})}^{n + 1})} = \frac{1}{3} \times \frac{{(\frac{2}{3})}^{n}}{(1 - {(\frac{2}{3})}^{n}) (1 - {(\frac{2}{3})}^{n + 1})}

\frac{x}{(1 - x) (1 - \frac{2}{3} x)} = 3 (\frac{1}{1 - x} - \frac{1}{1 - \frac{2}{3} x}) \Rightarrow

u_{n} = 3 (\frac{1}{1 - {(\frac{2}{3})}^{n}} - \frac{1}{1 - {(\frac{2}{3})}^{n + 1}}) = 3 (v_{n} - v_{n + 1}) with v_{n} = \frac{1}{1 - {(\frac{2}{3})}^{n}} \Rightarrow

\sum_{k = 1}^{n} u_{k} = 3 \sum_{k = 1}^{n} v_{k} - v_{k + 1} = 3 {v_{1} - v_{2} + v_{2} - v_{3} + \dots + v_{n} - v_{n + 1}}

= 3 (v_{1} - v_{n + 1}) = 3 (\frac{1}{1 - \frac{2}{3}} - \frac{1}{1 - {(\frac{2}{3})}^{n + 1}}) \Rightarrow

\lim_{n \to + \infty} \sum_{k = 1}^{n} u_{k} = 3 (3 - 1) = 6