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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 1 \sum \infty \frac{1}{3 n - 1}

Apply Series Comparison Test:

diverges

= 2 diverges

= diverges

n = 2 \sum \infty \frac{- 1}{ln ( n )}

n = 3 \sum \infty \frac{7}{n} - \frac{7}{n + 2}

k = 0 \sum \infty \frac{k}{4 k + 1}

n = 0 \sum \infty \frac{n}{8 n ^{6} + 1}

k = 1 \sum \infty 9 (\frac{2}{3})^{k}

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