sum from n=1 to infinity of (2/9)^n

Lösung

n = 1 \sum \infty (\frac{2}{9})^{n}

\frac{2}{7}

Dezimale

0.28571 \dots

Schritte zur Lösung

n = 1 \sum \infty (\frac{2}{9})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (\frac{2}{9})^{n} - (\frac{2}{9})^{0}

n = 0 \sum \infty (\frac{2}{9})^{n} = \frac{9}{7}

= \frac{9}{7} - (\frac{2}{9})^{0}

Vereinfache

\frac{9}{7} - (\frac{2}{9})^{0} : \frac{2}{7}

= \frac{2}{7}

n = 1 \sum \infty n e^{- 16 n}

n = 1 \sum \infty (- 0.005)^{n}

n = 1 \sum \infty \frac{2}{n ln ( 9 n )}

n = 1 \sum \infty \frac{n !}{11 6 ^{n}}

n = 1 \sum \infty \frac{1}{5 ^{n} - 2}