sum from n=1 to infinity of (2/13)^n

Lösung

n = 1 \sum \infty (\frac{2}{13})^{n}

\frac{2}{11}

Dezimale

0.18181 \dots

Schritte zur Lösung

n = 1 \sum \infty (\frac{2}{13})^{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}{13})^{n} - (\frac{2}{13})^{0}

n = 0 \sum \infty (\frac{2}{13})^{n} = \frac{13}{11}

= \frac{13}{11} - (\frac{2}{13})^{0}

Vereinfache

\frac{13}{11} - (\frac{2}{13})^{0} : \frac{2}{11}

= \frac{2}{11}

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

n = 0 \sum \infty (4 + n^{2})^{- 1}

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

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

n = 1 \sum \infty \frac{2 ^{n}}{n ^{4}}