What is the sum from n=0 to infinity of n/(3n^2+1) ?

The sum from n=0 to infinity of n/(3n^2+1) is diverges

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sum from n=0 to infinity of n/(3n^2+1)

n = 0 \sum \infty \frac{n}{3 n ^{2} + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{3 n ^{2} + 1}

Apply Series Integral Test:

diverges

= diverges

j = 1 \sum \infty 3^{- 3 j}

n = 1 \sum \infty \frac{3 n}{3 n + 1}

n = 0 \sum \infty 7^{n}

n = 1 \sum \infty \frac{( - \frac{1}{5} ) ^{n}}{n}

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

What is the sum from n=0 to infinity of n/(3n^2+1) ?
The sum from n=0 to infinity of n/(3n^2+1) is diverges