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

Lösung

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

1

Schritte zur Lösung

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

Teleskopreihen-Test anwenden:

1

= 1

k = 1 \sum \infty k^{2}

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

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

n = 1 \sum \infty \frac{ln ( n )}{9 n}

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