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 Comparison Test:

diverges

= diverges

f (x) = 6 x^{l n (x)}

- \frac{5}{x ^{3}}

f (x) = 3 x + 5

\int cos^{2} (wt - k x) d x

\int \frac{1}{x ^{2} - 4} d x

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