sum from n=2 to infinity of 3/(5^{n-1)}

解答

n = 2 \sum \infty \frac{3}{5 ^{n - 1}}

收敛

求解步骤

n = 2 \sum \infty \frac{3}{5 ^{n - 1}}

使用常数相乘法则:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot n = 2 \sum \infty \frac{1}{5 ^{n - 1}}

使用级数比值审敛法:

收敛

= 3 收敛

= 收敛

n = 1 \sum \infty \frac{3}{1 + n ^{2}}

n = 1 \sum \infty 2^{(- 1)^{n}}

n = 0 \sum \infty \frac{1 ^{n}}{n + 1}

n = 2 \sum \infty \frac{n !}{n ^{n}}

k = 4 \sum \infty \frac{1}{2 k ^{2} + k}