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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

\int 4 sec^{2} (x) d x

\frac{d}{d x} (x e)

\int \frac{8}{sec ( 2 x ) - 1} d x

\frac{\partial}{\partial x} (sin (x y) z^{2})

y = \frac{- 4 x}{x ^{2} + 1}, (0, 0)

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