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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

n = 0 \sum \infty ln (n + 1)

n = 0 \sum \infty \frac{1}{9 + n}

n = 0 \sum \infty \frac{n}{( n + 1 ) !}

n = 1 \sum \infty \frac{4}{n ^{2} - 1}

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