sum from k=0 to infinity of (0.1)^k

Lösung

k = 0 \sum \infty (0.1)^{k}

1.11111 \dots

Schritte zur Lösung

k = 0 \sum \infty 0. 1^{k}

Geometrischen Test der Reihe anwenden:

1.11111 \dots

= 1.11111 \dots

n = 0 \sum \infty \frac{2}{7 - 3 n}

n = 0 \sum \infty \frac{n}{n}

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

n = 0 \sum \infty 5 \cdot (\frac{2}{3})^{n}

n = 1 \sum \infty \frac{1}{n ^{2 - 1}}