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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\frac{\partial}{\partial x} (y^{5} cos (6 x) \cdot 6)

\frac{d}{d x} (\frac{1 + ln ( x )}{1 - ln ( x )})

\int \frac{4}{4 - x ^{2}} d x

- \frac{6 x}{x ^{2} + 1}, at x = 1

f (x) = 9 x^{2}

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