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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

\frac{d y}{d x} x^{2} y = y - 7

\int_{- 1}^{3} x^{2} - 2 x - 3 d x

\int \frac{6}{x ln ( 3 x )} d x

n = 1 \sum \infty \frac{5}{n n}

\int (2 t^{2} + sec^{2} (t)) d t

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