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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

\int x^{3} \cdot sin (3 x) d x

\int_{0}^{3} (5 x - x^{2}) - (2 x) d x

\frac{d ^{2}}{d x ^{2}} (- \frac{2 x ^{2}}{5 y ^{2}})

\frac{d y}{d x} = e^{6 y + 10 x}

\frac{d}{d x} (2 x - \frac{16}{x ^{2}})

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