What is the sum from n=1 to infinity of-1/(3(3n+2)) ?

The sum from n=1 to infinity of-1/(3(3n+2)) is diverges

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sum from n=1 to infinity of-1/(3(3n+2))

n = 1 \sum \infty - \frac{1}{3 ( 3 n + 2 )}

diverges

Solution steps

n = 1 \sum \infty - \frac{1}{3 ( 3 n + 2 )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - \frac{1}{3} \cdot n = 1 \sum \infty \frac{1}{3 n + 2}

Apply Series Integral Test:

diverges

= - \frac{1}{3} diverges

= diverges

\int \frac{1}{y ln ( y )} d y

\frac{\partial}{\partial v} (u + 2 v)

\int_{1}^{2} \frac{v ^{6} + 4 v ^{8}}{v ^{5}} d v

y^{^{''}} + 4 y^{^{'}} - 16 y = 0

\frac{\partial}{\partial y} (3 x^{2} (1 + ln (y)))

What is the sum from n=1 to infinity of-1/(3(3n+2)) ?
The sum from n=1 to infinity of-1/(3(3n+2)) is diverges