What is the sum from n=1 to infinity of (-1)^nsec(n) ?

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test

= diverges

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}}

What is the sum from n=1 to infinity of (-1)^nsec(n) ?
The sum from n=1 to infinity of (-1)^nsec(n) is diverges