What is the sum from n=1 to infinity of tan(3/(k^2)) ?

The sum from n=1 to infinity of tan(3/(k^2)) is diverges

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sum from n=1 to infinity of tan(3/(k^2))

n = 1 \sum \infty tan (\frac{3}{k ^{2}})

diverges

Solution steps

k = 1 \sum \infty tan (\frac{3}{n ^{2}})

Every infinite sum of a non-zero constant diverges

= diverges

n = 0 \sum \infty \frac{n}{3 n + 2}

n = 0 \sum \infty \frac{4}{7 - n}

n = 1 \sum \infty \frac{7 ^{n}}{n + 1}

n = 0 \sum \infty 2^{n} - n

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

What is the sum from n=1 to infinity of tan(3/(k^2)) ?
The sum from n=1 to infinity of tan(3/(k^2)) is diverges