sum from n=1 to infinity of (-1)^nsec(n)

Lösung

n = 1 \sum \infty (- 1)^{n} sec (n)

divergiert

Schritte zur Lösung

n = 1 \sum \infty (- 1)^{n} sec (n)

Seriendivergenztest anwenden

= divergiert

n = 1 \sum \infty \frac{1}{12 n + 11}

n = 3 \sum \infty (\frac{1}{4})^{n}

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

n = 0 \sum \infty 3 (\frac{2}{6})^{n + 2}

k = 1 \sum \infty \frac{k !}{10 4 ^{k}}