sum from n=1 to infinity of 2/(n(n+2))

解答

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

\frac{3}{2}

十进制

1.5

求解步骤

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

使用常数相乘法则:

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

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

将

\frac{1}{n ( n + 2 )}

用部份分式展开:

\frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

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

展开裂项级数:

\frac{3}{4}

化简

2 \cdot \frac{3}{4} : \frac{3}{2}

= \frac{3}{2}

\int \frac{sin ( x )}{cos ^{7} ( x )} d x

d er i v a t i v e 3 x^{2} e^{x} + e^{x} (x^{3} + 1)

\int - 4 x d x

\frac{d y}{d x} = \frac{y ( - 2 x + 4 )}{( x ^{2} + 1 ) ( x - 1 ) ^{2}}

d er i v a t i v e 3 arccos (\frac{x}{2})