sum from n=5 to infinity of 3(0.9)^{n+1}

解答

n = 5 \sum \infty 3 (0.9)^{n + 1}

收敛

求解步骤

n = 5 \sum \infty 3 \cdot 0. 9^{n + 1}

使用常数相乘法则:

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

= 3 \cdot n = 5 \sum \infty 0. 9^{n + 1}

化简

0. 9^{n + 1} : 0. 9^{n} \cdot 0.9

= 3 \cdot n = 5 \sum \infty 0. 9^{n} \cdot 0.9

使用常数相乘法则:

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

= 3 \cdot 0.9 \cdot n = 5 \sum \infty 0. 9^{n}

使用级数比值审敛法:

收敛

= 3 \cdot 0.9 收敛

= 收敛

x \to 0 lim (x - ln (x))

y^{^{'}} + y^{5} x + 9 y = 0

x \to - \frac{π}{2} lim (tan (x))

d er i v a t i v e x + 5 x

\int \frac{1}{2 e ^{2 x}} d x