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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

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