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

Lösung

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

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}}

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= \frac{5}{4} - \frac{1}{5 ^{0}}

Vereinfache

\frac{5}{4} - \frac{1}{5 ^{0}} : \frac{1}{4}

= \frac{1}{4}

\frac{1}{x} \cdot \frac{d y}{d x} + 2 y - e^{- x^{2}} = 0, y (1) = \frac{2}{e}

\frac{\partial}{\partial x} (lo g_{e} (x))

x \to - 2 + lim (f (x))

\int a e^{- a x} d x

(x^{3} cos (x))^{^{'}}