de

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Rechnen >

sum from n=1 to infinity of e^{-7n}

Lösung

n = 1 \sum \infty e^{- 7 n}

Lösung

konvergiert

Schritte zur Lösung

n = 1 \sum \infty e^{- 7 n}

Reihenverhältnis-Test anwenden:

konvergiert

= konvergiert

Beliebte Beispiele

sum from n=1 to infinity of 6e^{-n}

n = 1 \sum \infty 6 e^{- n}

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

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

sum from k=5 to infinity of 3(1/3)^{3k}

k = 5 \sum \infty 3 (\frac{1}{3})^{3 k}

sum from n=2 to infinity of (1/5)^{2n}

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

sum from k=0 to infinity of (0.3)^k

k = 0 \sum \infty (0.3)^{k}