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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 2 \sum \infty (\frac{3}{7})^{4 n}

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

n = 1 \sum \infty \frac{6 n}{4 n + 1}

n = 1 \sum \infty \frac{2}{5 n ^{2} + 5 n}

n = 0 \sum \infty (\frac{5}{12})^{n}

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