sum from k=4 to infinity of 1/(k^2-5k+6)

Lösung

k = 4 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

1

Schritte zur Lösung

k = 4 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

Teleskopreihen-Test anwenden:

1

= 1

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

n = 1 \sum \infty \frac{e ^{4 n}}{n !}

x = 1 \sum \infty x

k = 2 \sum \infty \frac{1}{k + 1} - \frac{1}{k}

n = 0 \sum \infty \frac{5}{2 ^{n - 1}}