sum from n=1 to infinity of 1/(9^n)

Lösung

n = 1 \sum \infty \frac{1}{9 ^{n}}

\frac{1}{8}

Dezimale

0.125

Schritte zur Lösung

n = 1 \sum \infty \frac{1}{9 ^{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}{9 ^{n}} - \frac{1}{9 ^{0}}

n = 0 \sum \infty \frac{1}{9 ^{n}} = \frac{9}{8}

= \frac{9}{8} - \frac{1}{9 ^{0}}

Vereinfache

\frac{9}{8} - \frac{1}{9 ^{0}} : \frac{1}{8}

= \frac{1}{8}

\frac{d}{d x} (\frac{x - 1}{x + 1})

x \to 4 lim (\frac{∣ 3 - x ∣}{3 - x})

\int_{0}^{\frac{π}{4}} cos^{2} (x) d x

\frac{\partial}{\partial y} (4 e^{y})

4 y^{^{''}} + 7 y^{^{'}} - 2 y = 0