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

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

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

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

diverges

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{tan ( n )} = tan^{- 1} (n)

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

Apply Series Divergence Test:

diverges

= diverges

n = 2 \sum \infty \frac{1}{n \cdot ( n - 1 )}

n = 0 \sum \infty \frac{2 n ^{2}}{n !}

k = 1 \sum \infty k e^{- 2 k^{2}}

n = 1 \sum \infty (\frac{7}{9})^{n - 1}

n = 3 \sum \infty \frac{4 ^{n}}{3 ^{n}}

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