sum from n=0 to infinity of sec(1/(n^2))

Lösung

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

divergiert

Schritte zur Lösung

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

Seriendivergenztest anwenden:

divergiert

= divergiert

n = 0 \sum \infty n^{2} e^{n^{3}}

n = 1 \sum \infty \frac{3}{8 ^{n} + 1}

k = 1 \sum \infty 9 k e^{- k}

n = 1 \sum \infty \frac{cos ( nπ )}{ln ( 3 n )}

n = 0 \sum \infty (1 + 7 n)^{\frac{1}{2}}