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

Lösung

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

divergiert

Schritte zur Lösung

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

Reihenintegraltest anwenden:

divergiert

= divergiert

\frac{\partial}{\partial y} (x sin (y) sin (z))

f (z) = \frac{1}{z}

\frac{d}{d x} (\frac{4 x}{x ^{2} + 4})

\int \frac{x ^{3}}{9 - x ^{2}} d x

\frac{\partial}{\partial y} (4 x^{2} + y^{2})