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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{4}{7 - n}

n = 1 \sum \infty \frac{7 ^{n}}{n + 1}

n = 0 \sum \infty 2^{n} - n

n = 1 \sum \infty (cos (3))^{n}

n = 0 \sum \infty \frac{1}{5} (\frac{4}{5})^{n}

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