sum from n=1 to infinity of n/(10n-3)

Lösung

n = 1 \sum \infty \frac{n}{10 n - 3}

divergiert

Schritte zur Lösung

n = 1 \sum \infty \frac{n}{10 n - 3}

Seriendivergenztest anwenden:

divergiert

= divergiert

n = 0 \sum \infty 5 (\frac{5}{2})^{n}

n = 0 \sum \infty (\frac{1}{100})^{n}

n = 0 \sum \infty \frac{9}{2 ^{n}}

n = 1 \sum \infty \frac{n ^{4}}{8 ^{n}}

n = 1 \sum \infty \frac{n}{n ^{2} + 16}