sum from b=0 to infinity of 9(2/3)^b

解答

b = 0 \sum \infty 9 (\frac{2}{3})^{b}

27

求解步骤

b = 0 \sum \infty 9 (\frac{2}{3})^{b}

使用常数相乘法则:

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

= 9 \cdot b = 0 \sum \infty (\frac{2}{3})^{b}

使用几何级数审敛法:

3

= 9 \cdot 3

化简

= 27

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

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

n = 1 \sum \infty n e^{- 15 n}

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

n = 1 \sum \infty \frac{1}{n ^{\frac{9}{8}}}