sum from n=1 to infinity of 1/(16n^2-8n-3)

Lösung

n = 1 \sum \infty \frac{1}{16 n ^{2} - 8 n - 3}

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

n = 1 \sum \infty \frac{1}{16 n ^{2} - 8 n - 3}

Teleskopreihen-Test anwenden:

\frac{1}{4}

= \frac{1}{4}

n = 0 \sum \infty \frac{2}{( 1 - n ) ^{3}}

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

n = 1 \sum \infty \frac{n !}{11 4 ^{n}}

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 1 \sum \infty n + 1