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

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

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

n = 1 \sum \infty \frac{n - 1}{3 n - 1}

diverges

Solution steps

n = 1 \sum \infty \frac{n - 1}{3 n - 1}

Apply Series Divergence Test:

diverges

= diverges

y^{^{''}} + 1500 y = 320

f (x) = ln (18)

\int cos (x) sin (sin (x)) d x

y = 6 cos (π x), y = 8 x^{2} - 2, - 0.5, 0.5

\int \frac{1}{2 + 2 x} d x

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