sum from i=1 to infinity of 1/(4i^2-1)

Lösung

i = 1 \sum \infty \frac{1}{4 i ^{2} - 1}

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

i = 1 \sum \infty \frac{1}{4 i ^{2} - 1}

Teleskopreihen-Test anwenden:

\frac{1}{2}

= \frac{1}{2}

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

n = 1 \sum \infty cos (6)

n = 1 \sum \infty \frac{n}{5 n ^{2} - 4}

n = 0 \sum \infty tanh (n)

n = 0 \sum \infty 8