sum from n=2 to infinity of (2/5)^n

Lösung

n = 2 \sum \infty (\frac{2}{5})^{n}

\frac{4}{15}

Dezimale

0.26666 \dots

Schritte zur Lösung

n = 2 \sum \infty (\frac{2}{5})^{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}{5})^{n} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1}

n = 0 \sum \infty (\frac{2}{5})^{n} = \frac{5}{3}

= \frac{5}{3} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1}

Vereinfache

\frac{5}{3} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1} : \frac{4}{15}

= \frac{4}{15}

n = 1 \sum \infty \frac{4 ^{n}}{n ^{7}}

k = 1 \sum \infty \frac{1}{k ^{7}}

n = 0 \sum \infty \frac{5 ^{n}}{7 ^{n}}

n = 0 \sum \infty n (0. 8^{n})

n = 0 \sum \infty - \frac{4}{e ^{n}}