sum from k=2 to infinity of 1/(k+1)-1/k

Lösung

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

- \frac{1}{2}

Dezimale

- 0.5

Schritte zur Lösung

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

Multipliziere aus

\frac{1}{k + 1} - \frac{1}{k} : - \frac{1}{k ^{2} + k}

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

Wende die Regel der konstanten Multiplikation:

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

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

Teleskopreihen-Test anwenden:

\frac{1}{2}

= - \frac{1}{2}

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

n = 0 \sum \infty \frac{n}{1 - 4 n}

n = 16 \sum \infty \frac{1}{ln ( n )}

n = 1 \sum \infty \frac{1}{9 n}

k = 2 \sum \infty (\frac{3}{5})^{3 k}