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

Lösung

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

\frac{3}{4}

Dezimale

0.75

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Erweitere die Teleskopreihe:

\frac{3}{4}

= \frac{3}{4}

n = 1 \sum \infty n^{- 0.8}

n = 2 \sum \infty \frac{4}{n ( n + 2 )}

k = 1 \sum \infty \frac{k}{e ^{k}}

n = 0 \sum \infty (1 + \frac{3}{n})^{n}

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