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

Lösung

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

- 2

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

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

Wende Exponentenregel an:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{1 ^{n}}{2 ^{n}} = (\frac{1}{2})^{n}

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

Geometrischen Test der Reihe anwenden:

2

= - 2

n = 0 \sum \infty cos (\frac{8}{n})

n = 0 \sum \infty (- 0.75)^{n}

n = 0 \sum \infty \frac{1}{5 - n}

n = 1 \sum \infty e^{n} + e^{- n}

n = 1 \sum \infty \frac{n}{7 n ^{2} + 1}