sum from n=3 to infinity of 1/(n^2-n)

Lösung

n = 3 \sum \infty \frac{1}{n ^{2} - n}

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

n = 3 \sum \infty \frac{1}{n ^{2} - n}

Teleskopreihen-Test anwenden:

\frac{1}{2}

= \frac{1}{2}

n = 0 \sum \infty \frac{n}{9 n + 1}

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

n = 0 \sum \infty \frac{3 ^{n}}{7 ^{n}}

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

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