What is the sum from n=0 to infinity of n/(9n^2+1) ?

The sum from n=0 to infinity of n/(9n^2+1) is diverges

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sum from n=0 to infinity of n/(9n^2+1)

n = 0 \sum \infty \frac{n}{9 n ^{2} + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{9 n ^{2} + 1}

Apply Series Integral Test:

diverges

= diverges

k = 0 \sum \infty (\frac{3}{5})^{k}

n = 1 \sum \infty 5 (\frac{1}{π})^{n}

n = 1 \sum \infty \frac{6}{n + 5}

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

n = 0 \sum \infty \frac{π}{2}

What is the sum from n=0 to infinity of n/(9n^2+1) ?
The sum from n=0 to infinity of n/(9n^2+1) is diverges