What is the sum from n=1 to infinity of (9n)/(4n+9) ?

The sum from n=1 to infinity of (9n)/(4n+9) is diverges

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sum from n=1 to infinity of (9n)/(4n+9)

n = 1 \sum \infty \frac{9 n}{4 n + 9}

diverges

Solution steps

n = 1 \sum \infty \frac{9 n}{4 n + 9}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{n}{4 n + 9}

Apply Series Divergence Test:

diverges

= 9 diverges

= diverges

\frac{d}{d x} (\frac{2}{1 - 3 x})

x \to 0 lim (\frac{x}{sin ( 3 x )})

(- 1, 2), (- 4, - 1)

y^{^{'}} + 4 y = cos (4 t)

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

What is the sum from n=1 to infinity of (9n)/(4n+9) ?
The sum from n=1 to infinity of (9n)/(4n+9) is diverges