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

解答

k = 1 \sum \infty 3 (- \frac{1}{5})^{3 k}

收敛

求解步骤

k = 1 \sum \infty 3 (- \frac{1}{5})^{3 k}

使用常数相乘法则:

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

= 3 \cdot k = 1 \sum \infty (- \frac{1}{5})^{3 k}

使用级数比值审敛法:

收敛

= 3 收敛

= 收敛

n = 0 \sum \infty - \frac{3}{( n ^{2} - 4 )}

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

n = 0 \sum \infty \frac{1}{n ^{\frac{5}{2}}}

n = 0 \sum \infty \frac{n}{e ^{7 n}}

n = 4 \sum \infty \frac{5 ^{n}}{1 4 ^{n}}