sum from n=0 to infinity of (3.5)^{-n}

Lösung

n = 0 \sum \infty (3.5)^{- n}

1.4

Schritte zur Lösung

n = 0 \sum \infty 3. 5^{- n}

Vereinfache

3. 5^{- n} : \frac{1}{3. 5 ^{n}}

= n = 0 \sum \infty \frac{1}{3. 5 ^{n}}

Geometrischen Test der Reihe anwenden:

1.4

= 1.4

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

n = 1 \sum \infty (\frac{π}{n})^{n} \cdot n!

n = 10 \sum \infty n

n = 0 \sum \infty n^{2} e^{- 3 n}

n = 2 \sum \infty (- 0.45)^{n}