What is the sum from n=1 to infinity of 2/(3n+9) ?

The sum from n=1 to infinity of 2/(3n+9) is diverges

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sum from n=1 to infinity of 2/(3n+9)

n = 1 \sum \infty \frac{2}{3 n + 9}

diverges

Solution steps

n = 1 \sum \infty \frac{2}{3 n + 9}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{3 n + 9}

Apply Series Limit Comparison Test:

diverges

= 2 diverges

= diverges

\int_{l n (x)}^{1} sin^{3} (u) d u

y = x^{3} - 12 x^{2} + 32 x, - x^{3} + 12 x^{2} - 32 x

\int e^{\frac{x}{y}} d x

x \to 2 lim (sin (\frac{2}{x - 2}))

\int (\frac{x + 3}{2 ^{x}} - 3) d x

What is the sum from n=1 to infinity of 2/(3n+9) ?
The sum from n=1 to infinity of 2/(3n+9) is diverges