sum from n=3 to infinity of 4/(n^2-n)

Lösung

n = 3 \sum \infty \frac{4}{n ^{2} - n}

2

Schritte zur Lösung

n = 3 \sum \infty \frac{4}{n ^{2} - n}

Wende die Regel der konstanten Multiplikation:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 3 \sum \infty \frac{1}{n ^{2} - n}

Teleskopreihen-Test anwenden:

\frac{1}{2}

= 4 \cdot \frac{1}{2}

Vereinfache

4 \cdot \frac{1}{2} : 2

= 2

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

n = 1 \sum \infty \frac{1}{n ^{\frac{4}{5}}}

n = 1 \sum \infty \frac{1}{9 n ^{2} + 23}

m = 2 \sum \infty \frac{6}{7 ^{m}}

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