sum from n=1 to infinity of (-1)^{n-1}

解答

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

发散

求解步骤

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

使用级数发散判别法:

发散

= 发散

\int \frac{3 x + 1}{( x - 2 ) ( x + 3 ) ^{2}} d x

\frac{\partial}{\partial x} (\frac{( y + z )}{x})

\frac{d}{d x} (k (sec (x) + tan (x)))

\int (1 + \frac{9}{4} x)^{\frac{1}{2}} d x

\frac{d y}{d x} = 1 + y