What is the sum from n=1 to infinity of (3n)/(4n+5) ?

The sum from n=1 to infinity of (3n)/(4n+5) is diverges

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sum from n=1 to infinity of (3n)/(4n+5)

n = 1 \sum \infty \frac{3 n}{4 n + 5}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n}{4 n + 5}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n}{4 n + 5}

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

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

(4 x^{3} + 4 x y - 1) d x = (1 - 2 x^{2} - 2 y) d y

\frac{d}{d x} (∣ x ∣^{∣ x ∣})

x \frac{d y}{d x} + y = - 9 x y^{2}

\frac{\partial}{\partial x} (\frac{8}{x ^{2}})

What is the sum from n=1 to infinity of (3n)/(4n+5) ?
The sum from n=1 to infinity of (3n)/(4n+5) is diverges