sum from n=1 to infinity of 1/(n^2+3n+2)

Lösung

n = 1 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

n = 1 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

Teleskopreihen-Test anwenden:

\frac{1}{2}

= \frac{1}{2}

\int \frac{( x + 1 )}{( x ^{2} + 2 x )} d x

\int \frac{6 x ^{3} - 12 x ^{2} + 8}{x ^{2} - 2 x} d x

t a y l or e^{x + 2}

(4 + x) y^{^{'}} = 7 y

x \to 0 lim (3 x^{2} + 2 x)