sum from n=0 to infinity of e^{-0.06n}

解答

n = 0 \sum \infty e^{- 0.06 n}

收敛

求解步骤

n = 0 \sum \infty e^{- 0.06 n}

n \to \infty lim (n (\frac{e ^{- 0.06 n}}{e ^{- 0.06 (n + 1)}} - 1)) = \infty

L > 1,

由拉比判别法

= 收敛

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{\frac{t}{2}}

\frac{d}{d x} (csc^{7} (2 x^{π} + π^{3}))

a re a sin^{2} (x), sin^{4} (x), [0, π]

\int (4 x + 1) x - 5 d x

\frac{d y}{d x} = 3 cos (7 x)