sum from n=0 to infinity of-(1/2)^n

Lösung

n = 0 \sum \infty - (\frac{1}{2})^{n}

- 2

Schritte zur Lösung

n = 0 \sum \infty - (\frac{1}{2})^{n}

Wende die Regel der konstanten Multiplikation:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 0 \sum \infty (\frac{1}{2})^{n}

Geometrischen Test der Reihe anwenden:

2

= - 2

j = 1 \sum \infty 2^{- 4 j}

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

n = 1 \sum \infty n! (- e)^{- 2 n}

n = 0 \sum \infty e^{\frac{- n ^{6}}{8}}

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