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

解答

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

收敛

求解步骤

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

使用常数相乘法则:

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

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

使用级数积分审敛法:

收敛

= 2 收敛

= 收敛

\frac{d P}{d t} = P - P^{2}

\frac{d y}{d x} = \frac{3}{x ^{2} y - 8 y}

\frac{\partial}{\partial x} (R (x)^{2} + x^{2} - x)

d er i v a t i v e f (0) = (x^{4} + 3 x) (4 x^{5} + 4 x - 2)

d er i v a t i v e - \frac{2}{x ^{2}}