sum from n=1 to infinity of (13/2)^n

Lösung

n = 1 \sum \infty (\frac{13}{2})^{n}

divergiert

Schritte zur Lösung

n = 1 \sum \infty (\frac{13}{2})^{n}

n \to \infty lim (n (\frac{( \frac{13}{2} ) ^{n}}{( \frac{13}{2} ) ^{n + 1}} - 1)) = - \infty

L < 1,

aufgrund des Raabe Tests

= divergiert

k = 1 \sum \infty \frac{4 k}{e ^{k}}

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

n = 1 \sum \infty \frac{31}{n !}

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

n = 0 \sum \infty 5^{n} 4^{- n}