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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

\frac{d y}{d x} = \frac{( x + y )}{3 x + 3 y - 4}

(2 x + 3 y) d x + (3 x + 2 y) d y = 0

9 - 6 x

\frac{d}{d x} (ln (x + y) - ln (x - y))

\frac{\partial}{\partial x} (cos (x) - x cos (y))

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