sum from k=2 to infinity of (1/3)^k

Lösung

k = 2 \sum \infty (\frac{1}{3})^{k}

\frac{1}{6}

Dezimale

0.16666 \dots

Schritte zur Lösung

k = 2 \sum \infty (\frac{1}{3})^{k}

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

= k = 0 \sum \infty (\frac{1}{3})^{k} - (\frac{1}{3})^{0} - (\frac{1}{3})^{1}

k = 0 \sum \infty (\frac{1}{3})^{k} = \frac{3}{2}

= \frac{3}{2} - (\frac{1}{3})^{0} - (\frac{1}{3})^{1}

Vereinfache

\frac{3}{2} - (\frac{1}{3})^{0} - (\frac{1}{3})^{1} : \frac{1}{6}

= \frac{1}{6}

d er i v a t i v e y = arcsin (x) + 1 - x^{2}

\frac{\partial}{\partial v} (uv - 1)

\frac{d}{d x} (4 x + \frac{64}{x})

d er i v a t i v e f (x) = 3 x - 9

x \to 3 lim (\frac{x ^{2} - 2 x + 1}{2})