sum from n=3 to infinity of (2/3)^n

Lösung

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

\frac{8}{9}

Dezimale

0.88888 \dots

Schritte zur Lösung

n = 3 \sum \infty (\frac{2}{3})^{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}{3})^{n} - (\frac{2}{3})^{0} - (\frac{2}{3})^{1} - (\frac{2}{3})^{2}

n = 0 \sum \infty (\frac{2}{3})^{n} = 3

= 3 - (\frac{2}{3})^{0} - (\frac{2}{3})^{1} - (\frac{2}{3})^{2}

Vereinfache

3 - (\frac{2}{3})^{0} - (\frac{2}{3})^{1} - (\frac{2}{3})^{2} : \frac{8}{9}

= \frac{8}{9}

n = 1 \sum \infty \frac{1}{n ^{\frac{4}{5}}}

n = 1 \sum \infty \frac{1}{9 n ^{2} + 23}

m = 2 \sum \infty \frac{6}{7 ^{m}}

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

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