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

Lösung

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

- 1

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

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

Teleskopreihen-Test anwenden:

\frac{1}{2}

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

Vereinfache

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

= - 1

n = 1 \sum \infty \frac{1}{n ^{0.2}}

n = 1 \sum \infty (0.9)^{n - 1}

n = 2 \sum \infty n sin (\frac{10}{n})

n = 1 \sum \infty (\frac{3}{π})^{n}

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