sum from k=7 to infinity of 1/(k^2-5k+6)

Lösung

k = 7 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

k = 7 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

Teleskopreihen-Test anwenden:

\frac{1}{4}

= \frac{1}{4}

k = 0 \sum \infty (\frac{4}{7})^{k}

x = 1 \sum \infty \frac{x + 2}{x + 1}

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

n = 0 \sum \infty \frac{cos ( πn )}{5}

n = 11 \sum \infty \frac{17}{3 ^{n}}