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

Lösung

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

divergiert

Schritte zur Lösung

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

Seriendivergenztest anwenden:

divergiert

= divergiert

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

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

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

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

n = 2 \sum \infty ln (1)