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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

\frac{\partial}{\partial x} (ln (x + y) + x \cdot cos (y))

\frac{d}{d x} (ln (x e^{- 9 x}))

\frac{\partial}{\partial x} (\frac{c - b}{a - x})

\frac{d}{d x} (m x^{n} - n x^{m})

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