sum from n=0 to infinity of (n^2)/(5+2n)

Lösung

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

divergiert

Schritte zur Lösung

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

Seriendivergenztest anwenden:

divergiert

= divergiert

n = 1 \sum \infty (sin (2))^{n}

i = 1 \sum \infty 5 \cdot 0. 2^{i - 1}

n = 1 \sum \infty \frac{3}{5 ^{n + 2}}

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

n = 0 \sum \infty 5^{- 2 n + 1} 3^{n}