sum from n=0 to infinity of n/(10n^2+1)

Lösung

n = 0 \sum \infty \frac{n}{10 n ^{2} + 1}

divergiert

Schritte zur Lösung

n = 0 \sum \infty \frac{n}{10 n ^{2} + 1}

Reihenintegraltest anwenden:

divergiert

= divergiert

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

n = 1 \sum \infty 7 (0.1)^{n - 1}

n = 1 \sum \infty \frac{8}{6 n ^{2} + 6 n}

n = 4 \sum \infty ln (1 + \frac{1}{n})

n = 2 \sum \infty \frac{7}{3 ^{n}}