What is the sum from n=0 to infinity of (n+4)/(3n+2) ?

The sum from n=0 to infinity of (n+4)/(3n+2) is diverges

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sum from n=0 to infinity of (n+4)/(3n+2)

n = 0 \sum \infty \frac{n + 4}{3 n + 2}

diverges

Solution steps

n = 0 \sum \infty \frac{n + 4}{3 n + 2}

Apply Series Divergence Test:

diverges

= diverges

\int (z - 3)^{n} d z

\int_{- 1}^{1} (7 x^{3} - 2 x^{2} + 5 x - 4) d x

\frac{d y}{d x} = \frac{2 x + 1}{2 y}

f (x) = \frac{1}{x + 3}

(ln (x))^{^{'}}

What is the sum from n=0 to infinity of (n+4)/(3n+2) ?
The sum from n=0 to infinity of (n+4)/(3n+2) is diverges