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

n = 2 \sum \infty (\frac{2}{5})^{n}

n = 1 \sum \infty \frac{4 ^{n}}{n ^{7}}

k = 1 \sum \infty \frac{1}{k ^{7}}

n = 0 \sum \infty \frac{5 ^{n}}{7 ^{n}}

n = 0 \sum \infty n (0. 8^{n})

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