What is the sum from n=0 to infinity of 1/(3n+1) ?

The sum from n=0 to infinity of 1/(3n+1) is diverges

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sum from n=0 to infinity of 1/(3n+1)

n = 0 \sum \infty \frac{1}{3 n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{3 n + 1}

Apply Series Integral Test:

diverges

= diverges

(\frac{2}{x})^{^{'}}

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

\frac{d y}{d x} = y (y^{2} - 1)

y = x^{2} + 2 x - 4

y = x^{3} + 3 x^{2}

What is the sum from n=0 to infinity of 1/(3n+1) ?
The sum from n=0 to infinity of 1/(3n+1) is diverges