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

解答

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

收敛

求解步骤

n = 1 \sum \infty \frac{( - 1 ) ^{n}}{5 \cdot 2 ^{n + 2}}

使用常数相乘法则:

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

= \frac{1}{5} \cdot n = 1 \sum \infty \frac{( - 1 ) ^{n}}{2 ^{n + 2}}

化简

2^{n + 2} : 2^{n} \cdot 2^{2}

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

使用常数相乘法则:

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

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

化简

= \frac{1}{5} \cdot \frac{1}{4} \cdot n = 1 \sum \infty \frac{( - 1 ) ^{n}}{2 ^{n}}

使用交错级数审敛法:

收敛

= \frac{1}{5} \cdot \frac{1}{4} 收敛

= 收敛

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