sum from n=0 to infinity of 4/(9^n)

Lösung

n = 0 \sum \infty \frac{4}{9 ^{n}}

\frac{9}{2}

Dezimale

4.5

Schritte zur Lösung

n = 0 \sum \infty \frac{4}{9 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{9 ^{n}}

Geometrischen Test der Reihe anwenden:

\frac{9}{8}

= 4 \cdot \frac{9}{8}

Vereinfache

4 \cdot \frac{9}{8} : \frac{9}{2}

= \frac{9}{2}

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

n = 1 \sum \infty n (\frac{8}{9})^{n}

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 2}

n = 1 \sum \infty 6 ln (\frac{n}{n + 1})

k = 0 \sum \infty \frac{k}{2 k + 1}