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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty (- \frac{1}{8})^{n}

n = 1 \sum \infty 8 ln (\frac{n}{n + 1})

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

n = 0 \sum \infty \frac{n}{n !}

n = 4 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

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