อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Influenza_Mathematics
3. จงหาค่าของ $$\sum_{k = 1}^{\infty}\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})} $$
|
$\begin {array}{rl}
\displaystyle \sum_{k = 1}^{\infty}\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}&=\lim_{n\to\infty}\sum_{k = 1}^{n}\left(\frac{2^k}{3^k-2^k}-\frac{2^{k+1}}{3^{k+1}-2^{k+1}}\right)\\
&=\lim_{n\to\infty}\left(\frac{2^1}{3^1-2^1}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}\right)\\
&=2-\lim_{n\to\infty}\frac{1}{(\frac{3}{2})^{n+1}-1}\\
&=2\end {array}$