What is the sum from n=0 to infinity of sec(1/(n^2)) ?

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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}}

What is the sum from n=0 to infinity of sec(1/(n^2)) ?
The sum from n=0 to infinity of sec(1/(n^2)) is diverges