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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

cos^{2} (x) + 1

\int cos^{2} (x) + 1 d x

\int_{0}^{\frac{1 - u}{4}} 2 (1 - y) d y

\int_{0}^{1} e^{x} (4 - e^{x}) d x

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

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