sum from k=0 to infinity of (1/3)^k

Lösung

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

\frac{3}{2}

Dezimale

1.5

Schritte zur Lösung

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

Geometrischen Test der Reihe anwenden:

\frac{3}{2}

= \frac{3}{2}

n = 0 \sum \infty \frac{2}{3 - n}

n = 1 \sum \infty - 3 (\frac{8}{7})^{n - 1}

n = 0 \sum \infty \frac{1 + ( - 1 ) ^{n}}{n}

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

n = 1 \sum \infty \frac{4}{e ^{n}}