数列(2^n-1)/(3^n)的极限给出具体求解,高手都一眼看出来了,可是考试不能写一眼看出来,(2^n-1)/(3^n)=(2^n)/(3^n)-1/(3^n),然后怎么求解,
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![数列(2^n-1)/(3^n)的极限给出具体求解,高手都一眼看出来了,可是考试不能写一眼看出来,(2^n-1)/(3^n)=(2^n)/(3^n)-1/(3^n),然后怎么求解,](/uploads/image/z/5604476-68-6.jpg?t=%E6%95%B0%E5%88%97%282%5En-1%29%2F%283%5En%29%E7%9A%84%E6%9E%81%E9%99%90%E7%BB%99%E5%87%BA%E5%85%B7%E4%BD%93%E6%B1%82%E8%A7%A3%2C%E9%AB%98%E6%89%8B%E9%83%BD%E4%B8%80%E7%9C%BC%E7%9C%8B%E5%87%BA%E6%9D%A5%E4%BA%86%2C%E5%8F%AF%E6%98%AF%E8%80%83%E8%AF%95%E4%B8%8D%E8%83%BD%E5%86%99%E4%B8%80%E7%9C%BC%E7%9C%8B%E5%87%BA%E6%9D%A5%2C%282%5En-1%29%2F%283%5En%29%3D%282%5En%29%2F%283%5En%29-1%2F%283%5En%29%2C%E7%84%B6%E5%90%8E%E6%80%8E%E4%B9%88%E6%B1%82%E8%A7%A3%2C)
数列(2^n-1)/(3^n)的极限给出具体求解,高手都一眼看出来了,可是考试不能写一眼看出来,(2^n-1)/(3^n)=(2^n)/(3^n)-1/(3^n),然后怎么求解,
数列(2^n-1)/(3^n)的极限
给出具体求解,高手都一眼看出来了,可是考试不能写一眼看出来,
(2^n-1)/(3^n)=(2^n)/(3^n)-1/(3^n),然后怎么求解,
数列(2^n-1)/(3^n)的极限给出具体求解,高手都一眼看出来了,可是考试不能写一眼看出来,(2^n-1)/(3^n)=(2^n)/(3^n)-1/(3^n),然后怎么求解,
[2^n-1]/3^n = (2^n/3^n) - 1/3^n = (2/3)^n - (1/3)^n.
n->无穷大时,(2/3)^n -> 0, (1/3)^n -> 0, 所以,[2^n - 1]/3^n = (2/3)^n - (1/3)^n -> 0-0 = 0.
若分开后,极限都存在,就可以分开求.
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