sum from n=0 to infinity of (e)^{-n}

Lösung

n = 0 \sum \infty (e)^{- n}

\frac{e}{e - 1}

Dezimale

1.58197 \dots

Schritte zur Lösung

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

Vereinfache

e^{- n} : \frac{1}{e ^{n}}

= n = 0 \sum \infty \frac{1}{e ^{n}}

Geometrischen Test der Reihe anwenden:

\frac{e}{e - 1}

= \frac{e}{e - 1}

n = 0 \sum \infty (\frac{4}{25})^{n}

n = 2 \sum \infty (\frac{3}{4})^{2 n}

k = 1 \sum \infty \frac{k}{500 k + 5}

n = 8 \sum \infty \frac{2 n + 1}{n ^{n}}

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