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

Lösung

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

divergiert

Schritte zur Lösung

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

Reihenintegraltest anwenden:

divergiert

= divergiert

n = 0 \sum \infty \frac{3 n}{n + 2}

n = 1 \sum \infty (cos (9))^{n}

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

n = 1 \sum \infty n^{4}

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